Find parent of leaf nodes in binary tree
The problem is to find the parents of all leaf nodes in a binary tree. A binary tree is a tree data structure in which each node can have at most two children, referred to as the left child and the right child. A leaf node is a node with no children, meaning it is the end of a branch in the tree.
In this context, we want to find and display the parents of all the leaf nodes in the given binary tree.
Pseudocode
leafParent(node, parent)
if node is not null
if node is a leaf node
if parent is null
print "NULL"
else
print parent.data
else
leafParent(node.left, node)
leafParent(node.right, node)
Algorithm Explanation
The leafParent
function finds the parents of leaf nodes in the binary tree using recursion. It takes two
parameters: node
and parent
. The variable node
represents the current node
being visited, while parent
represents the parent node of the current node.
The algorithm starts by checking if node
is not null. If node
is null, it means that there
are no more nodes to explore, and the function returns. Otherwise, it proceeds to check if the node
is
a leaf node, which means it has no left and right children.
If the node
is a leaf node, the algorithm prints the data of its parent. If the parent
is
null, it means the current node is the root of the tree, and there is no parent for the root. In this case, the
algorithm prints "NULL" to indicate that there is no parent for the root node. Otherwise, it prints the data of the
parent node.
If the node
is not a leaf node, the algorithm recursively calls itself on the left and right children of
the current node, passing the current node as the parent.
Code Solution
Here given code implementation process.

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The time complexity of the provided algorithm is O(N)
, where N
is the number of nodes in
the binary tree. This is because the algorithm visits each node exactly once in a depthfirst manner, and the number
of nodes visited is proportional to the number of nodes in the tree.
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