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by Sumit Chourasia | Oct 09, 2020 | Category :coding | Tags : algorithm binary-tree data-structure easy leetcode tree

Cousins in Binary Tree - Tree - Easy - LeetCode

Cousins in Binary Tree - Tree - Easy - LeetCode

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

 

Example 1:


Input: root = [1,2,3,4], x = 4, y = 3
Output: false
Example 2:


Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true
Example 3:

 

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false
 

Constraints:

The number of nodes in the tree will be between 2 and 100.
Each node has a unique integer value from 1 to 100.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public bool IsCousins(TreeNode root, int x, int y) {
        if(root == null){
            return false;
        }
        
        var queue = new Queue<TreeNode>();
        queue.Enqueue(root);
        while(queue.Count()>0){
            int count = queue.Count();
            bool foundx = false;
            bool foundy = false;
            while(count-->0){
                var node = queue.Dequeue();
                if(node.val==x){
                    foundx=true;
                }
                if(node.val==y){
                    foundy=true;
                }         
                
                if(node.left != null && node.right != null){
                    if(node.left.val ==x && node.right.val == y){
                        return false;
                    }
                    if(node.right.val==x && node.left.val==y){
                        return false;
                    }
                }
                
                if(node.left != null){
                    queue.Enqueue(node.left);
                }
                if(node.right != null){
                    queue.Enqueue(node.right);
                }
                            
            }
            if(foundx && foundy)
            {
                return true;
            }
        }
        
        return false;
    }
}

Time Complexity: O(n)

Space Complexity: O(height)