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.

- If right child of given node exists, then the right child is postorder predecessor.
- If right child does not exist and given node is left child of its parent, then its sibling is its postorder predecessor.
- 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"` `, ` ` ` `nodeToCheck->key); ` ` ` `} ` ` ` ` ` `return` `0; ` `} ` |

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*filter_none*

**Output:**

Postorder predecessor of 19 is 14

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

Auxiliary Space : O(1)

## Recommended Posts:

- Preorder predecessor of a Node in Binary Tree
- Level Order Predecessor of a node in Binary Tree
- Replace each node in binary tree with the sum of its inorder predecessor and successor
- Postorder successor of a Node in Binary Tree
- Find n-th node in Postorder traversal of a Binary Tree
- Construct a Binary Search Tree from given postorder
- Search a node in Binary Tree
- Insert a node in Binary Search Tree Iteratively
- Find the node with minimum value in a Binary Search Tree
- Find the node with maximum value in a Binary Search Tree
- Construct a Binary Tree from Postorder and Inorder
- Minimum swap required to convert binary tree to binary search tree
- Postorder traversal of Binary Tree without recursion and without stack
- Construct Full Binary Tree from given preorder and postorder traversals
- Complexity of different operations in Binary tree, Binary Search Tree and AVL tree

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