# Postorder successor of a Node in Binary Tree

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

```Examples: Consider the following binary tree
20
/      \
10       26
/  \     /   \
4     18  24    27
/  \
14   19
/  \
13  15

Postorder traversal of given tree is 4, 13, 15, 14,
19, 18, 10, 24, 27, 26, 20.

Input :  24
Output : 27

Input : 4
Output : 13
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 given node is root then postorder successor is NULL, since root is the last node print in a postorder traversal
2. If given node is right child of parent or right child of parent is NULL, then parent is postorder successor.
3. If given node is left child of parent and right child of parent is not NULL, then postorder successor is the leftmost node of parent’s right subtree

## C++

 `// CPP program to find postorder successor of ` `// given node. ` `#include ` `using` `namespace` `std; ` `  `  `struct` `Node { ` `    ``struct` `Node *left, *right, *parent; ` `    ``int` `value; ` `}; ` `  `  `// Utility function to create a new node with ` `// given value. ` `struct` `Node* newNode(``int` `value) ` `{ ` `    ``Node* temp = ``new` `Node; ` `    ``temp->left = temp->right = temp->parent = NULL; ` `    ``temp->value = value; ` `    ``return` `temp; ` `} ` `  `  `Node* postorderSuccessor(Node* root, Node* n) ` `{ ` `    ``// Root has no successor in postorder ` `    ``// traversal ` `    ``if` `(n == root) ` `        ``return` `NULL; ` `  `  `    ``// If given node is right child of its ` `    ``// parent or parent's right is empty, then  ` `    ``// parent is postorder successor. ` `    ``Node* parent = n->parent; ` `    ``if` `(parent->right == NULL || parent->right == n) ` `        ``return` `parent; ` `  `  `    ``// In all other cases, find the leftmost  ` `    ``// child in right substree of parent. ` `    ``Node* curr = parent->right; ` `    ``while` `(curr->left != NULL) ` `        ``curr = curr->left; ` `  `  `    ``return` `curr; ` `} ` `  `  `// Driver code ` `int` `main() ` `{ ` `    ``struct` `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; ` `  `  `    ``struct` `Node* res = postorderSuccessor(root, root->left->right->right); ` `    ``if` `(res)  ` `        ``printf``(``"Postorder successor of %d is %d\n"``, ` `               ``root->left->right->right->value, res->value);     ` `    ``else` `        ``printf``(``"Postorder successor of %d is NULL\n"``, ` `               ``root->left->right->right->value);     ` `  `  `    ``return` `0; ` `}`

## Java

 `// Java program to find postorder successor of  ` `// given node.  ` `import` `java.util.*; ` `class` `GfG { ` ` `  `static` `class` `Node {  ` `    ``Node left, right, parent;  ` `    ``int` `value;  ` `} ` ` `  `// Utility function to create a new node with  ` `// given value.  ` `static` `Node newNode(``int` `value)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.left = ``null``; ` `    ``temp.right = ``null``; ` `    ``temp.parent = ``null``;  ` `    ``temp.value = value;  ` `    ``return` `temp;  ` `}  ` ` `  `static` `Node postorderSuccessor(Node root, Node n)  ` `{  ` `    ``// Root has no successor in postorder  ` `    ``// traversal  ` `    ``if` `(n == root)  ` `        ``return` `null``;  ` ` `  `    ``// If given node is right child of its  ` `    ``// parent or parent's right is empty, then  ` `    ``// parent is postorder successor.  ` `    ``Node parent = n.parent;  ` `    ``if` `(parent.right == ``null` `|| parent.right == n)  ` `        ``return` `parent;  ` ` `  `    ``// In all other cases, find the leftmost  ` `    ``// child in right substree of parent.  ` `    ``Node curr = parent.right;  ` `    ``while` `(curr.left != ``null``)  ` `        ``curr = curr.left;  ` ` `  `    ``return` `curr;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``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 = postorderSuccessor(root, root.left.right.right);  ` `    ``if` `(res != ``null``)  ` `        ``System.out.println(``"Postorder successor of "``+  ` `        ``root.left.right.right.value + ``" is "``+ res.value);  ` `    ``else` `        ``System.out.println(``"Postorder successor of "` `+  ` `        ``root.left.right.right.value + ``" is NULL"``); ` `}  ` `} `

## Python3

 `""" Python3 program to find postorder  ` `successor of a node in Binary Tree."""` ` `  `# A Binary Tree Node  ` `# Utility function to create a new tree node  ` `class` `newNode:  ` ` `  `    ``# Constructor to create a new node  ` `    ``def` `__init__(``self``, data):  ` `        ``self``.value ``=` `data  ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `        ``self``.parent``=``None` ` `  `def` `postorderSuccessor(root, n) : ` ` `  `    ``# Root has no successor in postorder  ` `    ``# traversal  ` `    ``if` `(n ``=``=` `root): ` `        ``return` `None` `     `  `    ``# If given node is right child of its  ` `    ``# parent or parent's right is empty,  ` `    ``# then parent is postorder successor.  ` `    ``parent ``=` `n.parent  ` `    ``if` `(parent.right ``=``=` `None` `or` `parent.right ``=``=` `n): ` `        ``return` `parent  ` `     `  `    ``# In all other cases, find the leftmost  ` `    ``# child in right substree of parent.  ` `    ``curr ``=` `parent.right  ` `    ``while` `(curr.left !``=` `None``): ` `        ``curr ``=` `curr.left  ` `     `  `    ``return` `curr ` `     `  `# 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 ``=` `postorderSuccessor(root, root.left.right.right)  ` ` `  `    ``if` `(res) :  ` `        ``print``(``"postorder successor of"``,  ` `               ``root.left.right.right.value,  ` `               ``"is"``, res.value)  ` `     `  `    ``else``: ` `        ``print``(``"postorder successor of"``,  ` `               ``root.left.right.right.value,  ` `                                 ``"is None"``) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

## C#

 `// C# program to find postorder successor of  ` `// given node. ` `using` `System; ` ` `  `class` `GfG  ` `{ ` ` `  `class` `Node  ` `{  ` `    ``public` `Node left, right, parent;  ` `    ``public` `int` `value;  ` `} ` ` `  `// Utility function to create   ` `// a new node with given value.  ` `static` `Node newNode(``int` `value)  ` `{  ` `    ``Node temp = ``new` `Node();  ` `    ``temp.left = ``null``; ` `    ``temp.right = ``null``; ` `    ``temp.parent = ``null``;  ` `    ``temp.value = value;  ` `    ``return` `temp;  ` `}  ` ` `  `static` `Node postorderSuccessor(Node root, Node n)  ` `{  ` `    ``// Root has no successor in   ` `    ``// postorder traversal  ` `    ``if` `(n == root)  ` `        ``return` `null``;  ` ` `  `    ``// If given node is right child of its  ` `    ``// parent or parent's right is empty, then  ` `    ``// parent is postorder successor.  ` `    ``Node parent = n.parent;  ` `    ``if` `(parent.right == ``null` `|| parent.right == n)  ` `        ``return` `parent;  ` ` `  `    ``// In all other cases, find the leftmost  ` `    ``// child in right substree of parent.  ` `    ``Node curr = parent.right;  ` `    ``while` `(curr.left != ``null``)  ` `        ``curr = curr.left;  ` ` `  `    ``return` `curr;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``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 = postorderSuccessor(root, root.left.right.right);  ` `    ``if` `(res != ``null``)  ` `        ``Console.WriteLine(``"Postorder successor of "``+  ` `        ``root.left.right.right.value + ``" is "``+ res.value);  ` `    ``else` `        ``Console.WriteLine(``"Postorder successor of "` `+  ` `        ``root.left.right.right.value + ``" is NULL"``); ` `}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```Postorder successor of 19 is 18
```

Time Complexity : O(h) where h is height of given Binary Tree
Auxiliary Space : O(1) since no use of arrays, stacks, queues.

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