Posted on by Kalkicode
Code Recursion

# Print path from root to a given node in a binary tree

The problem involves finding and printing the path from the root of a binary tree to a given node. This path represents the sequence of nodes that need to be traversed from the root to reach the desired node.

## Problem Statement

Given a binary tree and a target node, the task is to find and print the path from the root of the tree to the target node.

## Example Scenario

Consider the binary tree shown below:

``````
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1``````

For node 1, the path from the root is: [4, 9, 1].
For node 5, the path from the root is: [4, 7, 12, 5].

## Idea to Solve the Problem

To solve this problem, we can use a recursive approach to traverse the binary tree while maintaining a current path. At each step, we update the current path by adding the current node's value. When the target node is reached, we print the path.

## Pseudocode

``````void findRootToNodePath(TreeNode node, ArrayList path, int target)
{
if (node == null)
return;

if (node.data == target)
printPath(path);

findRootToNodePath(node.left, path, target);
findRootToNodePath(node.right, path, target);

path.remove(path.size() - 1);
}``````

## Algorithm Explanation

1. Implement a function `findRootToNodePath` that takes three arguments: `node` (current node), `path` (list of nodes in the path), and `target` (target node value).
2. Base case: If the current node is null, return.
3. Add the current node's value to the path.
4. If the current node's value is equal to the target value, print the path.
5. Recursively traverse the left and right subtrees while updating the path.
6. After processing the current node, remove it from the path to backtrack.

## Program Solution

``````import java.util.ArrayList;
/*
Java Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree
{
public TreeNode root;
public int count;
public BinaryTree()
{
// Set initial tree root to null
this.root = null;
this.count = 0;
}
// Display given path
public void printPath(ArrayList < Integer > path)
{
int i = 0;
// print path
while (i < path.size())
{
// print path node
System.out.print(" " + path.get(i));
i++;
}
System.out.print("\n");
}
// Find all root to given node path using recursion
public void findRootToNodePath(TreeNode point,
ArrayList < Integer > path, int node)
{
if (point == null)
{
return;
}
if (point.data == node)
{
this.count++;
printPath(path);
}
// Visit left and right subtree using recursion
findRootToNodePath(point.left, path, node);
findRootToNodePath(point.right, path, node);
// Remove last node in path
path.remove(path.size() - 1);
}
// Handles the request of finding root to given node path in tree
public void rootToNodePath(int node)
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
ArrayList < Integer > path = new ArrayList < Integer > ();
this.count = 0;
System.out.println(" \n Given node : " + node);
findRootToNodePath(this.root, path, node);
if (this.count == 0)
{
System.out.print("None");
}
}
}
public static void main(String[] args)
{
// Create new binary tree
BinaryTree tree = new BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(1);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(19);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(18);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(1);
// Case A
// All path from root to 1 node
tree.rootToNodePath(1);
// Case B
tree.rootToNodePath(5);
}
}``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````
``````// Include header file
#include <iostream>
#include <vector>
using namespace std;

/*
C++ Program
Print path from root to a given node in a binary tree
*/

// Binary Tree node
class TreeNode
{
public:
int data;
TreeNode *left;
TreeNode *right;
TreeNode(int data)
{
// Set node value
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
class BinaryTree
{
public:
TreeNode *root;
int count;
BinaryTree()
{
this->root = NULL;
this->count = 0;
}
// Display given path
void printPath(vector < int > path)
{
int i = 0;
// print path
while (i < path.size())
{
// print path node
cout << " " << path.at(i);
i++;
}
cout << "\n";
}
// Find all root to given node path using recursion
void findRootToNodePath(TreeNode *point,
vector < int > &path, int node)
{
if (point == NULL)
{
return;
}
path.push_back(point->data);
if (point->data == node)
{
this->count++;
this->printPath(path);
}
// Visit left and right subtree using recursion
this->findRootToNodePath(point->left, path, node);
this->findRootToNodePath(point->right, path, node);
// Remove last node in path
path.pop_back();
}
// Handles the request of finding root to given node path in tree
void rootToNodePath(int node)
{
if (this->root == NULL)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
vector < int > path;
this->count = 0;
cout << " \n Given node : " << node << endl;
this->findRootToNodePath(this->root, path, node);
if (this->count == 0)
{
cout << "None";
}
}
}
};
int main()
{
// Create new binary tree
BinaryTree *tree = new BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree->root = new TreeNode(4);
tree->root->left = new TreeNode(9);
tree->root->left->right = new TreeNode(1);
tree->root->left->right->left = new TreeNode(6);
tree->root->left->right->left->left = new TreeNode(19);
tree->root->left->right->right = new TreeNode(8);
tree->root->left->right->right->left = new TreeNode(1);
tree->root->left->right->right->left->right = new TreeNode(10);
tree->root->left->left = new TreeNode(2);
tree->root->right = new TreeNode(7);
tree->root->right->right = new TreeNode(12);
tree->root->right->right->right = new TreeNode(18);
tree->root->right->right->left = new TreeNode(5);
tree->root->right->right->left->right = new TreeNode(15);
tree->root->right->right->left->right->right = new TreeNode(1);
// Case A
// All path from root to 1 node
tree->rootToNodePath(1);
// Case B
tree->rootToNodePath(5);
return 0;
}``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````
``````// Include namespace system
using System;
using System.Collections.Generic;
/*
Csharp Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
public class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree
{
public TreeNode root;
public int count;
public BinaryTree()
{
// Set initial tree root to null
this.root = null;
this.count = 0;
}
// Display given path
public void printPath(List < int > path)
{
int i = 0;
// print path
while (i < path.Count)
{
// print path node
Console.Write(" " + path[i]);
i++;
}
Console.Write("\n");
}
// Find all root to given node path using recursion
public void findRootToNodePath(TreeNode point,
List < int > path, int node)
{
if (point == null)
{
return;
}
if (point.data == node)
{
this.count++;
this.printPath(path);
}
// Visit left and right subtree using recursion
this.findRootToNodePath(point.left, path, node);
this.findRootToNodePath(point.right, path, node);
// Remove last node in path
path.RemoveAt(path.Count - 1);
}
// Handles the request of finding root to given node path in tree
public void rootToNodePath(int node)
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
List < int > path = new List < int > ();
this.count = 0;
Console.WriteLine(" \n Given node : " + node);
this.findRootToNodePath(this.root, path, node);
if (this.count == 0)
{
Console.Write("None");
}
}
}
public static void Main(String[] args)
{
// Create new binary tree
BinaryTree tree = new BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(1);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(19);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(18);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(1);
// Case A
// All path from root to 1 node
tree.rootToNodePath(1);
// Case B
tree.rootToNodePath(5);
}
}``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````
``````<?php
/*
Php Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
class TreeNode
{
public \$data;
public \$left;
public \$right;
public	function __construct(\$data)
{
// Set node value
\$this->data = \$data;
\$this->left = NULL;
\$this->right = NULL;
}
}
class BinaryTree
{
public \$root;
public \$count;
public	function __construct()
{
\$this->root = NULL;
\$this->count = 0;
}
// Display given path
public	function printPath(\$path)
{
\$i = 0;
// print path
while (\$i < count(\$path))
{
// print path node
echo(" ".\$path[\$i]);
\$i++;
}
echo("\n");
}
// Find all root to given node path using recursion
public	function findRootToNodePath(\$point, &\$path, \$node)
{
if (\$point == NULL)
{
return;
}
\$path[] = \$point->data;
if (\$point->data == \$node)
{
\$this->count++;
\$this->printPath(\$path);
}
// Visit left and right subtree using recursion
\$this->findRootToNodePath(\$point->left, \$path, \$node);
\$this->findRootToNodePath(\$point->right, \$path, \$node);
// Remove last node in path
array_pop(\$path);
}
// Handles the request of finding root to given node path in tree
public	function rootToNodePath(\$node)
{
if (\$this->root == NULL)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
\$path = array();
\$this->count = 0;
echo(" \n Given node : ".\$node.
"\n");
\$this->findRootToNodePath(\$this->root, \$path, \$node);
if (\$this->count == 0)
{
echo("None");
}
}
}
}

function main()
{
// Create new binary tree
\$tree = new BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
\$tree->root = new TreeNode(4);
\$tree->root->left = new TreeNode(9);
\$tree->root->left->right = new TreeNode(1);
\$tree->root->left->right->left = new TreeNode(6);
\$tree->root->left->right->left->left = new TreeNode(19);
\$tree->root->left->right->right = new TreeNode(8);
\$tree->root->left->right->right->left = new TreeNode(1);
\$tree->root->left->right->right->left->right = new TreeNode(10);
\$tree->root->left->left = new TreeNode(2);
\$tree->root->right = new TreeNode(7);
\$tree->root->right->right = new TreeNode(12);
\$tree->root->right->right->right = new TreeNode(18);
\$tree->root->right->right->left = new TreeNode(5);
\$tree->root->right->right->left->right = new TreeNode(15);
\$tree->root->right->right->left->right->right = new TreeNode(1);
// Case A
// All path from root to 1 node
\$tree->rootToNodePath(1);
// Case B
\$tree->rootToNodePath(5);
}
main();``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````
``````/*
Node JS Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
class TreeNode
{
constructor(data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
constructor()
{
this.root = null;
this.count = 0;
}
// Display given path
printPath(path)
{
var i = 0;
// print path
while (i < path.length)
{
// print path node
process.stdout.write(" " + path[i]);
i++;
}
process.stdout.write("\n");
}
// Find all root to given node path using recursion
findRootToNodePath(point, path, node)
{
if (point == null)
{
return;
}
path.push(point.data);
if (point.data == node)
{
this.count++;
this.printPath(path);
}
// Visit left and right subtree using recursion
this.findRootToNodePath(point.left, path, node);
this.findRootToNodePath(point.right, path, node);
// Remove last node in path
path.pop();
}
// Handles the request of finding root to given node path in tree
rootToNodePath(node)
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
var path = [];
this.count = 0;
console.log(" \n Given node : " + node);
this.findRootToNodePath(this.root, path, node);
if (this.count == 0)
{
process.stdout.write("None");
}
}
}
}

function main()
{
// Create new binary tree
var tree = new BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(1);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(19);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(18);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(1);
// Case A
// All path from root to 1 node
tree.rootToNodePath(1);
// Case B
tree.rootToNodePath(5);
}
main();``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````
``````#    Python 3 Program
#    Print path from root to a given node in a binary tree

#  Binary Tree node
class TreeNode :
def __init__(self, data) :
#  Set node value
self.data = data
self.left = None
self.right = None

class BinaryTree :
def __init__(self) :
self.root = None
self.count = 0

#  Display given path
def printPath(self, path) :
i = 0
#  print path
while (i < len(path)) :
#  print path node
print(" ", path[i], end = "")
i += 1

print(end = "\n")

#  Find all root to given node path using recursion
def findRootToNodePath(self, point, path, node) :
if (point == None) :
return

path.append(point.data)
if (point.data == node) :
self.count += 1
self.printPath(path)

#  Visit left and right subtree using recursion
self.findRootToNodePath(point.left, path, node)
self.findRootToNodePath(point.right, path, node)
#  Remove last node in path
del path[len(path) - 1]

#  Handles the request of finding root to given node path in tree
def rootToNodePath(self, node) :
if (self.root == None) :
#  Empty Tree
return
else :
#  This is use to collect path
path = []
self.count = 0
print(" \n Given node : ", node)
self.findRootToNodePath(self.root, path, node)
if (self.count == 0) :
print("None", end = "")

def main() :
#  Create new binary tree
tree = BinaryTree()
#     4
#    /   \
#   9     7
#  / \     \
# 2   1     12
#    / \    / \
#   6   8  5   18
#  /   /    \
# 19  1      15
#      \      \
#       10     1
# -----------------
# Constructing binary tree
tree.root = TreeNode(4)
tree.root.left = TreeNode(9)
tree.root.left.right = TreeNode(1)
tree.root.left.right.left = TreeNode(6)
tree.root.left.right.left.left = TreeNode(19)
tree.root.left.right.right = TreeNode(8)
tree.root.left.right.right.left = TreeNode(1)
tree.root.left.right.right.left.right = TreeNode(10)
tree.root.left.left = TreeNode(2)
tree.root.right = TreeNode(7)
tree.root.right.right = TreeNode(12)
tree.root.right.right.right = TreeNode(18)
tree.root.right.right.left = TreeNode(5)
tree.root.right.right.left.right = TreeNode(15)
tree.root.right.right.left.right.right = TreeNode(1)
#  Case A
#  All path from root to 1 node
tree.rootToNodePath(1)
#  Case B
tree.rootToNodePath(5)

if __name__ == "__main__": main()``````

#### input

`````` Given node :  1
4  9  1
4  9  1  8  1
4  7  12  5  15  1

Given node :  5
4  7  12  5``````
``````#    Ruby Program
#    Print path from root to a given node in a binary tree

#  Binary Tree node
class TreeNode
# Define the accessor and reader of class TreeNode
attr_accessor :data, :left, :right
def initialize(data)
#  Set node value
self.data = data
self.left = nil
self.right = nil
end

end

class BinaryTree
# Define the accessor and reader of class BinaryTree
attr_accessor :root, :count
def initialize()
self.root = nil
self.count = 0
end

#  Display given path
def printPath(path)
i = 0
#  print path
while (i < path.length)
#  print path node
print(" ", path[i])
i += 1
end

print("\n")
end

#  Find all root to given node path using recursion
def findRootToNodePath(point, path, node)
if (point == nil)
return
end

path.push(point.data)
if (point.data == node)
self.count += 1
self.printPath(path)
end

#  Visit left and right subtree using recursion
self.findRootToNodePath(point.left, path, node)
self.findRootToNodePath(point.right, path, node)
#  Remove last node in path
path.delete_at(path.length - 1)
end

#  Handles the request of finding root to given node path in tree
def rootToNodePath(node)
if (self.root == nil)
#  Empty Tree
return
else

#  This is use to collect path
path = []
self.count = 0
print(" \n Given node : ", node, "\n")
self.findRootToNodePath(self.root, path, node)
if (self.count == 0)
print("None")
end

end

end

end

def main()
#  Create new binary tree
tree = BinaryTree.new()
#     4
#    /   \
#   9     7
#  / \     \
# 2   1     12
#    / \    / \
#   6   8  5   18
#  /   /    \
# 19  1      15
#      \      \
#       10     1
# -----------------
# Constructing binary tree
tree.root = TreeNode.new(4)
tree.root.left = TreeNode.new(9)
tree.root.left.right = TreeNode.new(1)
tree.root.left.right.left = TreeNode.new(6)
tree.root.left.right.left.left = TreeNode.new(19)
tree.root.left.right.right = TreeNode.new(8)
tree.root.left.right.right.left = TreeNode.new(1)
tree.root.left.right.right.left.right = TreeNode.new(10)
tree.root.left.left = TreeNode.new(2)
tree.root.right = TreeNode.new(7)
tree.root.right.right = TreeNode.new(12)
tree.root.right.right.right = TreeNode.new(18)
tree.root.right.right.left = TreeNode.new(5)
tree.root.right.right.left.right = TreeNode.new(15)
tree.root.right.right.left.right.right = TreeNode.new(1)
#  Case A
#  All path from root to 1 node
tree.rootToNodePath(1)
#  Case B
tree.rootToNodePath(5)
end

main()``````

#### input

``````
Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5
``````
``````import scala.collection.mutable._;
/*
Scala Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
class TreeNode(var data: Int,
var left: TreeNode,
var right: TreeNode)
{
def this(data: Int)
{
// Set node value
this(data,null,null);
}
}
class BinaryTree(var root: TreeNode,
var count: Int)
{
def this()
{
this(null,0);
}
// Display given path
def printPath(path: ArrayBuffer[Int]): Unit = {
var i: Int = 0;
// print path
while (i < path.size)
{
// print path node
print(" " + path(i));
i += 1;
}
print("\n");
}
// Find all root to given node path using recursion
def findRootToNodePath(point: TreeNode, path: ArrayBuffer[Int], node: Int): Unit = {
if (point == null)
{
return;
}
path += point.data;
if (point.data == node)
{
this.count += 1;
printPath(path);
}
// Visit left and right subtree using recursion
findRootToNodePath(point.left, path, node);
findRootToNodePath(point.right, path, node);
// Remove last node in path
path.remove(path.size - 1);
}
// Handles the request of finding root to given node path in tree
def rootToNodePath(node: Int): Unit = {
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
var path: ArrayBuffer[Int] = new ArrayBuffer[Int]();
this.count = 0;
println(" \n Given node : " + node);
findRootToNodePath(this.root, path, node);
if (this.count == 0)
{
print("None");
}
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
// Create new binary tree
var tree: BinaryTree = new BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(1);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(19);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(18);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(1);
// Case A
// All path from root to 1 node
tree.rootToNodePath(1);
// Case B
tree.rootToNodePath(5);
}
}``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````
``````import Foundation;
/*
Swift 4 Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode? ;
var right: TreeNode? ;
init(_ data: Int)
{
// Set node value
self.data = data;
self.left = nil;
self.right = nil;
}
}
class BinaryTree
{
var root: TreeNode? ;
var count: Int;
init()
{
self.root = nil;
self.count = 0;
}
// Display given path
func printPath(_ path: [Int])
{
var i = 0;
// print path
while (i < path.count)
{
// print path node
print(" ", path[i], terminator: "");
i += 1;
}
print(terminator: "\n");
}
// Find all root to given node path using recursion
func findRootToNodePath(_ point: TreeNode? , _ path : inout[Int], _ node: Int)
{
if (point == nil)
{
return;
}
path.append(point!.data);
if (point!.data == node)
{
self.count += 1;
self.printPath(path);
}
// Visit left and right subtree using recursion
self.findRootToNodePath(point!.left, &path, node);
self.findRootToNodePath(point!.right, &path, node);
// Remove last node in path
path.removeLast();
}
// Handles the request of finding root to given node path in tree
func rootToNodePath(_ node: Int)
{
if (self.root == nil)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
var path = [Int]();
self.count = 0;
print(" \n Given node : ", node);
self.findRootToNodePath(self.root, &path, node);
if (self.count == 0)
{
print("None", terminator: "");
}
}
}
}
func main()
{
// Create new binary tree
let tree = BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root!.left = TreeNode(9);
tree.root!.left!.right = TreeNode(1);
tree.root!.left!.right!.left = TreeNode(6);
tree.root!.left!.right!.left!.left = TreeNode(19);
tree.root!.left!.right!.right = TreeNode(8);
tree.root!.left!.right!.right!.left = TreeNode(1);
tree.root!.left!.right!.right!.left!.right = TreeNode(10);
tree.root!.left!.left = TreeNode(2);
tree.root!.right = TreeNode(7);
tree.root!.right!.right = TreeNode(12);
tree.root!.right!.right!.right = TreeNode(18);
tree.root!.right!.right!.left = TreeNode(5);
tree.root!.right!.right!.left!.right = TreeNode(15);
tree.root!.right!.right!.left!.right!.right = TreeNode(1);
// Case A
// All path from root to 1 node
tree.rootToNodePath(1);
// Case B
tree.rootToNodePath(5);
}
main();``````

#### input

`````` Given node :  1
4  9  1
4  9  1  8  1
4  7  12  5  15  1

Given node :  5
4  7  12  5``````
``````/*
Kotlin Program
Print path from root to a given node in a binary tree
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode ? ;
var right: TreeNode ? ;
constructor(data: Int)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
var root: TreeNode ? ;
var count: Int;
constructor()
{
this.root = null;
this.count = 0;
}
// Display given path
fun printPath(path: MutableList < Int > ): Unit
{
var i: Int = 0;
// print path
while (i < path.size)
{
// print path node
print(" " + path[i]);
i += 1;
}
print("\n");
}
// Find all root to given node path using recursion
fun findRootToNodePath(point: TreeNode ? ,
path : MutableList < Int >  , node : Int): Unit
{
if (point == null)
{
return;
}
if (point.data == node)
{
this.count += 1;
this.printPath(path);
}
// Visit left and right subtree using recursion
this.findRootToNodePath(point.left, path, node);
this.findRootToNodePath(point.right, path, node);
// Remove last node in path
path.removeAt(path.size - 1);
}
// Handles the request of finding root to given node path in tree
fun rootToNodePath(node: Int): Unit
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect path
var path: MutableList < Int > = mutableListOf < Int > ();
this.count = 0;
println(" \n Given node : " + node);
this.findRootToNodePath(this.root, path, node);
if (this.count == 0)
{
print("None");
}
}
}
}
fun main(args: Array < String > ): Unit
{
// Create new binary tree
val tree: BinaryTree = BinaryTree();
/*
4
/   \
9     7
/ \     \
2   1     12
/ \    / \
6   8  5   18
/   /    \
19  1      15
\      \
10     1
-----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root?.left = TreeNode(9);
tree.root?.left?.right = TreeNode(1);
tree.root?.left?.right?.left = TreeNode(6);
tree.root?.left?.right?.left?.left = TreeNode(19);
tree.root?.left?.right?.right = TreeNode(8);
tree.root?.left?.right?.right?.left = TreeNode(1);
tree.root?.left?.right?.right?.left?.right = TreeNode(10);
tree.root?.left?.left = TreeNode(2);
tree.root?.right = TreeNode(7);
tree.root?.right?.right = TreeNode(12);
tree.root?.right?.right?.right = TreeNode(18);
tree.root?.right?.right?.left = TreeNode(5);
tree.root?.right?.right?.left?.right = TreeNode(15);
tree.root?.right?.right?.left?.right?.right = TreeNode(1);
// Case A
// All path from root to 1 node
tree.rootToNodePath(1);
// Case B
tree.rootToNodePath(5);
}``````

#### input

`````` Given node : 1
4 9 1
4 9 1 8 1
4 7 12 5 15 1

Given node : 5
4 7 12 5``````

## Output Explanation

The code implements the algorithm and finds and prints the path from the root of the binary tree to the given target node. It demonstrates the result by printing the path for each test case.

## Time Complexity

The time complexity of this algorithm is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the traversal. The space complexity is O(H), where H is the height of the binary tree, due to the recursive call stack and the space used to store the current path.

## Comment

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