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Preorder Successor of a Node in Binary Tree
• Difficulty Level : Easy
• Last Updated : 19 Jun, 2019

Given a binary tree and a node in the binary tree, find preorder successor of the given node. It may be assumed that every node has parent link.
Examples:

Consider the following binary tree
20
/      \
10       26
/  \     /   \
4     18  24    27
/  \
14   19
/  \
13  15
Input :  4
Output : 18
Preorder traversal of given tree is 20, 10, 4,
18, 14, 13, 15, 19, 26, 24, 27.

Input :  19
Output : 26

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

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.

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

An efficient solution is based on below observations.

1. If left child of given node exists, then the left child is preorder successor.
2. If left child does not exist and given node is left child of its parent, then its sibling is its preorder successor.
3. If none of above conditions are satisfied (left child does not exist and given node is not left child of its parent), then we move up using parent pointers until one of the following happens.
• We reach root. In this case, preorder successor does not exist.
• Current node (one of the ancestors of given node) is left child of its parent, in this case preorder successor is sibling of current node.

## C++

 // CPP program to find preorder successor 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* preorderSuccessor(Node* root, Node* n) {     // If left child exists, then it is preorder     // successor.     if (n->left)         return n->left;        // If left child does not exist, then     // travel up (using parent pointers)     // until we reach a node which is left     // child of its parent.     Node *curr = n, *parent = curr->parent;     while (parent != NULL && parent->right == curr) {         curr = curr->parent;         parent = parent->parent;     }        // If we reached root, then the given     // node has no preorder successor     if (parent == NULL)         return NULL;        return parent->right; }    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 = preorderSuccessor(root, root->left->right->right);        if (res) {         printf("Preorder successor of %d is %d\n",                root->left->right->right->key, res->key);     }     else {         printf("Preorder successor of %d is NULL\n",                root->left->right->right->key);     }        return 0; }

## Java

 // Java program to find preorder successor of // a node in Binary Tree. class Solution {    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 preorderSuccessor(Node root, Node n) {     // If left child exists, then it is preorder     // successor.     if (n.left != null)         return n.left;        // If left child does not exist, then     // travel up (using parent pointers)     // until we reach a node which is left     // child of its parent.     Node curr = n, parent = curr.parent;     while (parent != null && parent.right == curr)      {         curr = curr.parent;         parent = parent.parent;     }        // If we reached root, then the given     // node has no preorder successor     if (parent == null)         return null;        return parent.right; }    // 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 = preorderSuccessor(root, root.left.right.right);        if (res != null)      {         System.out.printf("Preorder successor of %d is %d\n",             root.left.right.right.key, res.key);     }     else     {         System.out.printf("Preorder successor of %d is null\n",             root.left.right.right.key);     } } }    // This code is contributed by Arnab Kundu

## Python3

 """ Python3 program to find preorder 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.key = data          self.left = None         self.right = None         self.parent=None    def preorderSuccessor(root, n) :        # If left child exists, then it is     # preorder successor.      if (n.left) :         return n.left         # If left child does not exist, then      # travel up (using parent pointers)      # until we reach a node which is left      # child of its parent.      curr = n     parent = curr.parent      while (parent != None and             parent.right == curr):          curr = curr.parent          parent = parent.parent             # If we reached root, then the given      # node has no preorder successor      if (parent == None) :         return None        return parent.right        # 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 = preorderSuccessor(root, root.left.right.right)         if (res) :          print("Preorder successor of",                root.left.right.right.key, "is", res.key)             else:         print("Preorder successor of",                root.left.right.right.key,"is None")    # This code is contributed  # by SHUBHAMSINGH10

## C#

 // C# program to find preorder successor of // a node in Binary Tree. using System;        class GFG {    public 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 preorderSuccessor(Node root, Node n) {     // If left child exists, then it is preorder     // successor.     if (n.left != null)         return n.left;        // If left child does not exist, then     // travel up (using parent pointers)     // until we reach a node which is left     // child of its parent.     Node curr = n, parent = curr.parent;     while (parent != null && parent.right == curr)      {         curr = curr.parent;         parent = parent.parent;     }        // If we reached root, then the given     // node has no preorder successor     if (parent == null)         return null;        return parent.right; }    // 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 = preorderSuccessor(root, root.left.right.right);        if (res != null)      {         Console.Write("Preorder successor of {0} is {1}\n",             root.left.right.right.key, res.key);     }     else     {         Console.Write("Preorder successor of {0} is null\n",             root.left.right.right.key);     } } }    // This code is contributed by 29AjayKumar

Output:

Preorder successor of 19 is 26

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

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