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.

Examples: Consider the following binary tree
              20            
           /      \         
          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* res = postorderPredecessor(root, root->left->right->right);
  
    if (res) {
        printf("Postorder predecessor of %d is %d\n",
               root->left->right->right->key, res->key);
    }
    else {
        printf("Postorder predecessor of %d is NULL\n",
               root->left->right->right->key);
    }
  
    return 0;
}

Output:

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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