Given a binary tree and a node in the binary tree, find preorder successor of the given node. It may be assumed that every node has parent link.
Consider the following binary tree 20 / \ 10 26 / \ / \ 4 18 24 27 / \ 14 19 / \ 13 15 Input : 4 Output : 18 Preorder traversal of given tree is 20, 10, 4, 18, 14, 13, 15, 19, 26, 24, 27. Input : 19 Output : 26
A simple solution is to first store Preorder traversal of the given tree in an array then linearly search given node and print node next to it.
Time Complexity : O(n)
Auxiliary Space : O(n)
An efficient solution is based on below observations.
- If left child of given node exists, then the left child is preorder successor.
- If left child does not exist and given node is left child of its parent, then its sibling is its preorder successor.
- If none of above conditions are satisfied (left child does not exist and given node is not left child of its parent), then we move up using parent pointers until one of the following happens.
- We reach root. In this case, preorder successor does not exist.
- Current node (one of the ancestors of given node) is left child of its parent, in this case preorder successor is sibling of current node.
“”” Python3 program to find preorder
successor of a node in Binary Tree.”””
# A Binary Tree Node
# Utility function to create a new tree node
# Constructor to create a new node
def __init__(self, data):
self.key = data
self.left = None
self.right = None
def preorderSuccessor(root, n) :
# If left child exists, then it is
# preorder successor.
if (n.left) :
# If left child does not exist, then
# travel up (using parent pointers)
# until we reach a node which is left
# child of its parent.
curr = n
parent = curr.parent
while (parent != None and
parent.right == curr):
curr = curr.parent
parent = parent.parent
# If we reached root, then the given
# node has no preorder successor
if (parent == None) :
# Driver Code
if __name__ == ‘__main__’:
root = newNode(20)
root.parent = None
root.left = newNode(10)
root.left.parent = root
root.left.left = newNode(4)
root.left.left.parent = root.left
root.left.right = newNode(18)
root.left.right.parent = root.left
root.right = newNode(26)
root.right.parent = root
root.right.left = newNode(24)
root.right.left.parent = root.right
root.right.right = newNode(27)
root.right.right.parent = root.right
root.left.right.left = newNode(14)
root.left.right.left.parent = root.left.right
root.left.right.left.left = newNode(13)
root.left.right.left.left.parent = root.left.right.left
root.left.right.left.right = newNode(15)
root.left.right.left.right.parent = root.left.right.left
root.left.right.right = newNode(19)
root.left.right.right.parent = root.left.right
res = preorderSuccessor(root, root.left.right.right)
if (res) :
print(“Preorder successor of”,
root.left.right.right.key, “is”, res.key)
print(“Preorder successor of”,
# This code is contributed
# by SHUBHAMSINGH10
Preorder successor of 19 is 26
Time Complexity : O(h) where h is height of given Binary Tree
Auxiliary Space : O(1)
- Postorder successor of a Node in Binary Tree
- Inorder Successor of a node in Binary Tree
- Level Order Successor of a node in Binary Tree
- Replace each node in binary tree with the sum of its inorder predecessor and successor
- Preorder predecessor of a Node in Binary Tree
- Find n-th node in Preorder traversal of a Binary Tree
- Construct Full Binary Tree using its Preorder traversal and Preorder traversal of its mirror tree
- Modify a binary tree to get preorder traversal using right pointers only
- Calculate depth of a full Binary tree from Preorder
- Leaf nodes from Preorder of a Binary Search Tree (Using Recursion)
- Construct Full Binary Tree from given preorder and postorder traversals
- Check if a given array can represent Preorder Traversal of Binary Search Tree
- Sum of cousins of a given node in a Binary Tree
- K-th ancestor of a node in Binary Tree
- Search a node in Binary Tree
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.