Postorder predecessor of a Node in Binary Search Tree

Given a binary tree and a node in the binary tree, find Postorder predecessor of the given node.

Consider the following binary tree
           /      \         
          10       26       
         /  \     /   \     
       4     18  24    27   
            /  \
           14   19
          /  \
         13  15
Input :  4
Output : 10
Postorder traversal of given tree is 4, 13, 15, 
14, 19, 18, 10, 24, 27, 26, 20.

Input :  24
Output : 10

A simple solution is to first store Postorder 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.

  1. If right child of given node exists, then the right child is postorder predecessor.
  2. If right child does not exist and given node is left child of its parent, then its sibling is its postorder predecessor.
  3. If none of above conditions are satisfied (left child does not exist and given node is not right child of its parent), then we move up using parent pointers until one of the following happens.
    • We reach root. In this case, postorder predecessor does not exiss
    • Current node (one of the ancestors of given node) is right child of its parent, in this case postorder predecessor is sibling of current node.






// CPP program to find postorder predecessor of
// a node in Binary Tree.
#include <iostream>
using namespace std;
struct Node {
    struct Node *left, *right, *parent;
    int key;
Node* newNode(int key)
    Node* temp = new Node;
    temp->left = temp->right = temp->parent = NULL;
    temp->key = key;
    return temp;
Node* postorderPredecessor(Node* root, Node* n)
    // If right child exists, then it is postorder
    // predecessor.
    if (n->right)
        return n->right;
    // If right child does not exist, then
    // travel up (using parent pointers)
    // until we reach a node which is right
    // child of its parent.
    Node *curr = n, *parent = curr->parent;
    while (parent != NULL && parent->left == curr) {
        curr = curr->parent;
        parent = parent->parent;
    // If we reached root, then the given
    // node has no postorder predecessor
    if (parent == NULL)
        return NULL;
    return parent->left;
int main()
    Node* root = newNode(20);
    root->parent = NULL;
    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;
    Node* nodeToCheck = root->left->right;
    Node* res = postorderPredecessor(root, nodeToCheck);
    if (res) {
        printf("Postorder predecessor of %d is %d\n",
               nodeToCheck->key, res->key);
    else {
        printf("Postorder predecessor of %d is NULL\n",
    return 0;



“””Python3 program to find postorder
predecessor of given node.”””

# A Binary Tree Node
# Utility function to create a
# new tree node
class newNode:

# Constructor to create a newNode
def __init__(self, data):
self.key = data
self.left = None
self.right = self.parent = None

def postorderPredecessor(root, n):

# If right child exists, then it
# is postorder predecessor.
if (n.right) :
return n.right

# If right child does not exist, then
# travel up (using parent pointers)
# until we reach a node which is right
# child of its parent.
curr = n
parent = curr.parent
while (parent != None and
parent.left == curr):
curr = curr.parent
parent = parent.parent

# If we reached root, then the given
# node has no postorder predecessor
if (parent == None) :
return None

return parent.left

# 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

nodeToCheck = root.left.right

res = postorderPredecessor(root, nodeToCheck)

if (res) :
print(“Postorder predecessor of”,
nodeToCheck.key, “is”, res.key)

print(“Postorder predecessor of”,
nodeToCheck.key, “is None”)

# This code is contributed


Postorder predecessor of 19 is 14

Time Complexity: O(h) where h is height of given Binary Tree
Auxiliary Space: O(1)

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