# Preorder predecessor of a Node in Binary Tree

• Difficulty Level : Easy
• Last Updated : 18 Aug, 2022

Given a binary tree and a node in the binary tree, find Preorder 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
Preorder traversal of given tree is 20, 10, 4,
18, 14, 13, 15, 19, 26, 24, 27.

Input :  19
Output : 15```

A simple solution is to first store Preorder traversal of the given tree in an array then linearly search given node and print node next to it.

Complexity Analysis:

• Time Complexity : O(n)
• Auxiliary Space : O(n)

An efficient solution is based on below observations.

1. If the given node is root, then return NULL as preorder predecessor.
2. If node is the left child of its parent or left child of parent is NULL, then return parent as its preorder predecessor.
3. If node is the right child of its parent and left child of parent exists, then predecessor would be the rightmost node (max value) of the left subtree of parent.
4. If node is the right child of its parent and the parent has no left child, then predecessor would be the parent node (max value).

Implementation:

## C++

 `// CPP program to find preorder predecessor 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* preorderPredecessor(Node* root, Node* n)``{``    ``// Root has no predecessor in preorder``    ``// traversal``    ``if` `(n == root)``        ``return` `NULL;` `    ``// If given node is left child of its``    ``// parent or parent's left is empty, then``    ``// parent is Preorder Predecessor.``    ``Node* parent = n->parent;``    ``if` `(parent->left == NULL || parent->left == n)``        ``return` `parent;` `    ``// In all other cases, find the rightmost``    ``// child in left subtree of parent.``    ``Node* curr = parent->left;``    ``while` `(curr->right != NULL)``        ``curr = curr->right;` `    ``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 = preorderPredecessor(root, root->left->right->right);``    ``if` `(res)``        ``printf``(``"Preorder predecessor of %d is %d\n"``,``               ``root->left->right->right->value, res->value);   ``    ``else``        ``printf``(``"Preorder predecessor of %d is NULL\n"``,``               ``root->left->right->right->value);   ` `    ``return` `0;``}`

## Java

 `// Java program to find preorder predecessor of``// given node.``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 preorderPredecessor(Node root, Node n)``{``    ``// Root has no predecessor in preorder``    ``// traversal``    ``if` `(n == root)``        ``return` `null``;` `    ``// If given node is left child of its``    ``// parent or parent's left is empty, then``    ``// parent is Preorder Predecessor.``    ``Node parent = n.parent;``    ``if` `(parent.left == ``null` `|| parent.left == n)``        ``return` `parent;` `    ``// In all other cases, find the rightmost``    ``// child in left subtree of parent.``    ``Node curr = parent.left;``    ``while` `(curr.right != ``null``)``        ``curr = curr.right;` `    ``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 = preorderPredecessor(root, root.left.right.right);``    ``if` `(res != ``null``)``        ``System.out.println(``"Preorder predecessor of "` `+ root.left.right.right.value + ``" is "` `+ res.value);    ``    ``else``        ``System.out.println(``"Preorder predecessor of "` `+ root.left.right.right.value + ``" is null"``);` `}``}`

## Python3

 `"""Python3 program to find preorder``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``.value ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `self``.parent ``=` `None` `def` `preorderPredecessor(root, n):` `    ``# Root has no predecessor in preorder``    ``# traversal``    ``if` `(n ``=``=` `root) :``        ``return` `None` `    ``# If given node is left child of its``    ``# parent or parent's left is empty, then``    ``# parent is Preorder Predecessor.``    ``parent ``=` `n.parent``    ``if` `(parent.left ``=``=` `None` `or``        ``parent.left ``=``=` `n):``        ``return` `parent` `    ``# In all other cases, find the rightmost``    ``# child in left subtree of parent.``    ``curr ``=` `parent.left``    ``while` `(curr.right !``=` `None``) :``        ``curr ``=` `curr.right` `    ``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 ``=` `preorderPredecessor(root, root.left.right.right)``    ``if` `(res):``        ``print``(``"Preorder predecessor of"``,``               ``root.left.right.right.value, ``"is"``, res.value)    ``    ``else``:``        ``print``(``"Preorder predecessor of"``,``               ``root.left.right.right.value, ``"is None"``)` `# This code is contributed``# by SHUBHAMSINGH10`

## C#

 `// C# program to find preorder predecessor 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 preorderPredecessor(Node root, Node n)``    ``{``        ``// Root has no predecessor in preorder``        ``// traversal``        ``if` `(n == root)``            ``return` `null``;` `        ``// If given node is left child of its``        ``// parent or parent's left is empty, then``        ``// parent is Preorder Predecessor.``        ``Node parent = n.parent;``        ``if` `(parent.left == ``null` `|| parent.left == n)``            ``return` `parent;` `        ``// In all other cases, find the rightmost``        ``// child in left subtree of parent.``        ``Node curr = parent.left;``        ``while` `(curr.right != ``null``)``            ``curr = curr.right;` `        ``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 = preorderPredecessor(root, root.left.right.right);``        ``if` `(res != ``null``)``            ``Console.WriteLine(``"Preorder predecessor of "` `+``                               ``root.left.right.right.value +``                               ``" is "` `+ res.value);    ``        ``else``            ``Console.WriteLine(``"Preorder predecessor of "` `+``                                ``root.left.right.right.value +``                                ``" is null"``);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`Preorder predecessor of 19 is 15`

Complexity Analysis:

• 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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