Cousins in Binary Tree

Description

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

Note:

1. The number of nodes in the tree will be between 2 and 100.

2. Each node has a unique integer value from 1 to 100.

code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    Map<Integer, Integer> depth;
    Map<Integer, TreeNode> parent;

    public boolean isCousins(TreeNode root, int x, int y) {
        depth = new HashMap();
        parent = new HashMap();
        dfs(root, null);
        return (depth.get(x) == depth.get(y) && parent.get(x) != parent.get(y));
    }

    public void dfs(TreeNode node, TreeNode par) {
        if (node != null) {
            depth.put(node.val, par != null ? 1 + depth.get(par.val) : 0);
            parent.put(node.val, par);
            dfs(node.left, node);
            dfs(node.right, node);
        }
    }
}