Preorder Successor of a Node in Binary Tree

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

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++

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// CPP program to find preorder successor 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* 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;
}

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Java

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// 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

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Python3

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""" 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

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C#

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// 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 

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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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