Next Node on Its Right - Google Top Interview Questions


Problem Statement :


You are given a binary tree root containing unique values, and an integer target. 

Find the node with value target and return the node that's directly right of it on its level. 

If the target node doesn't exist or if it's already in the rightmost position, return null.

Constraints

0 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

Visualize

tree = [1, [2, [4, null, null], [5, null, null]], [3, null, [6, [7, null, null], null]]]

target = 5

Output



[6, [7, null, null], null]

Example 2

Input



tree = [1, [2, null, null], [3, null, null]]

target = 3

Output



null

Explanation

There's no node directly to the right of 3.



Example 3

Input



tree = [1, null, null]

target = 2

Output



null

Explanation

Node 2 doesn't exist.


Solution :



title-img 




                        Solution in C++ :

Tree* solve(Tree* tree, int target) {
    if (not tree) {
        return nullptr;
    }
    queue<Tree*> q;
    q.push(tree);
    Tree* cur;
    while (not q.empty()) {
        int s = q.size();
        while (s--) {
            cur = q.front();
            q.pop();

            if (cur->val == target) {
                if (s >= 1) {
                    Tree* now = q.front();
                    q.pop();
                    return now;
                }
            }

            if (cur->left) {
                q.push(cur->left);
            }
            if (cur->right) {
                q.push(cur->right);
            }
        }
    }
    return nullptr;
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    public Tree solve(Tree tree, int target) {
        Queue<Tree> q = new LinkedList();
        q.offer(tree);
        boolean trigger = false;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int x = 0; x < size; x++) {
                Tree temp = q.poll();
                if (trigger) {
                    return temp;
                }
                if (temp.val == target) {
                    trigger = true;
                }

                if (temp.left != null)
                    q.offer(temp.left);
                if (temp.right != null)
                    q.offer(temp.right);
            }
            if (trigger)
                return null;
        }
        return null;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, tree, target):
        q = deque()
        q.append(tree)
        while q:
            isInLevel = len(q)
            while isInLevel:
                temp = q.popleft()
                if temp.val == target:
                    if isInLevel > 1:
                        return q.popleft()
                    else:
                        return None
                if temp.left:
                    q.append(temp.left)
                if temp.right:
                    q.append(temp.right)
                isInLevel -= 1

        return None


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