# Detect next node at same level of a key in binary tree in python

Python program for Detect next node at same level of a key in binary tree. Here mentioned other language solution.

``````#  Python 3 program for
#  Find the next node of the key at the same
#  level in the 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.status = -1
self.result = None

#  Method which are find next node of same level
def findNextNode(self, temp, level, node) :
if (temp != None) :
if (self.status == -1 and temp.data == node) :
#  When found node value
self.status = level
elif (self.status == level and self.result == None) :
#  Next node of same level
self.result = temp

#  Visit to left subtree
self.findNextNode(temp.left, level + 1, node)
#  Visit to right subtree
self.findNextNode(temp.right, level + 1, node)

#  This are handling the request of to finding next
#  node in same level by given node
def nextNode(self, node) :
#  Reset the result indicator
self.status = -1
self.result = None
#  Find next node
self.findNextNode(self.root, 0, node)
print("\n Node : ", node)
print(" Result : ", end = "")
#  Test Case
if (self.status == -1) :
#  Case A
#  When node values are not exist
elif (self.result == None) :
#  Case B
print("None")
else :
#  Case C
#  When next node are exist
print(self.result.data)

def main() :
#  Create new tree
tree = BinaryTree()
#   Make A Binary Tree
#  ---------------------
#        1
#       / \
#      /   \
#     /     \
#    2       4
#   /       / \
#  /       /   \
#  3      6     5
#   \      \   /
#    7      8 11
#  Add node in Binary Tree
tree.root = TreeNode(1)
tree.root.left = TreeNode(2)
tree.root.left.left = TreeNode(3)
tree.root.left.left.right = TreeNode(7)
tree.root.right = TreeNode(4)
tree.root.right.right = TreeNode(5)
tree.root.right.right.left = TreeNode(11)
tree.root.right.left = TreeNode(6)
tree.root.right.left.right = TreeNode(8)
#  Testing
tree.nextNode(3)
tree.nextNode(7)
tree.nextNode(6)
tree.nextNode(5)
tree.nextNode(8)
tree.nextNode(1)

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

Output

`````` Node :  3
Result : 6

Node :  7
Result : 8

Node :  6
Result : 5

Node :  5
Result : None

Node :  8
Result : 11

Node :  1
Result : None``````

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