# Postorder predecessor of a Node in Binary Search Tree

Given a binary tree and a node in the binary tree, find the 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 the Postorder traversal of the given tree in an array then linearly search the given node and print the node next to it.
Time Complexity : O(n)
Auxiliary Space : O(n)
An efficient solution is based on the below observations.

1. If the right child of a given node exists, then the right child is the postorder predecessor.
2. If the right child does not exist and the given node is left child of its parent, then its sibling is its Postorder predecessor.
3. If none of the above conditions are satisfied (left child does not exist and given node is not the right child of its parent), then we move up using parent pointers until one of the following happens.
• We reach the root. In this case, the Postorder predecessor does not exist.
• The current node (one of the ancestors of the given node) is the right child of its parent, in this case, the postorder predecessor is the sibling of the current node.
 `// C++ program to find postorder predecessor of` `// a node in Binary Tree.` `#include ` `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;` `}`

 `// Java program to find postorder predecessor` `// of a node in Binary Tree.` `import` `java.util.*;`   `class` `GFG{`   `static` `class` `Node ` `{` `    ``Node left, right, parent;` `    ``int` `key;` `};`   `static` `Node newNode(``int` `key)` `{` `    ``Node temp = ``new` `Node();` `    ``temp.left = temp.right = temp.parent = ``null``;` `    ``temp.key = key;` `    ``return` `temp;` `}`   `static` `Node postorderPredecessor(Node root, Node n)` `{` `    `  `    ``// If right child exists, then it is ` `    ``// postorder predecessor.` `    ``if` `(n.right != ``null``)` `        ``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;` `}`   `// 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 nodeToCheck = root.left.right;`   `    ``Node res = postorderPredecessor(root, nodeToCheck);`   `    ``if` `(res != ``null``)` `    ``{` `        ``System.out.printf(``"Postorder predecessor "` `+ ` `                          ``"of %d is %d\n"``,` `                          ``nodeToCheck.key, res.key);` `    ``}` `    ``else` `    ``{` `        ``System.out.printf(``"Postorder predecessor "` `+ ` `                          ``"of %d is null\n"``,` `                          ``nodeToCheck.key);` `    ``}` `}` `}`   `// This code is contributed by Rajput-Ji`

 `"""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) ` `    `  `    ``else``:` `        ``print``(``"Postorder predecessor of"``, ` `               ``nodeToCheck.key, ``"is None"``)`   `# This code is contributed ` `# by SHUBHAMSINGH10`

 `// C# program to find postorder ` `// predecessor of a node ` `// in Binary Tree.` `using` `System;` `class` `GFG{`   `class` `Node ` `{` `  ``public` `Node left, right, parent;` `  ``public` `int` `key;` `};`   `static` `Node newNode(``int` `key)` `{` `  ``Node temp = ``new` `Node();` `  ``temp.left = temp.right = ` `              ``temp.parent = ``null``;` `  ``temp.key = key;` `  ``return` `temp;` `}`   `static` `Node postorderPredecessor(Node root, ` `                                 ``Node n)` `{    ` `  ``// If right child exists, ` `  ``// then it is postorder ` `  ``// predecessor.` `  ``if` `(n.right != ``null``)` `    ``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;` `}`   `// 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 nodeToCheck = root.left.right;`   `  ``Node res = postorderPredecessor(root, nodeToCheck);`   `  ``if` `(res != ``null``)` `  ``{` `    ``Console.Write(``"Postorder predecessor "` `+ ` `                  ``"of {0} is {1}\n"``,` `                  ``nodeToCheck.key, res.key);` `  ``}` `  ``else` `  ``{` `    ``Console.Write(``"Postorder predecessor "` `+ ` `                  ``"of {0} is null\n"``,` `                  ``nodeToCheck.key);` `  ``}` `}` `}`   `// This code is contributed by shikhasingrajput`

Output:
```Postorder predecessor of 18 is 19

```

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

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