Tags: "leetcode", "binary-tree", "dfs", access_time 2-min read

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Cousins in Binary Tree

Created: May 7, 2020 by [lek-tin]

Last updated: May 7, 2020

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:

cousins in binary tree example 1

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

Example 2:

cousins in binary tree example 2

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

Example 3:

cousins in binary tree 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.

Solution

Java

/**
 * 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 {
    boolean shared_parent = false;
    int x_level = -1;
    int y_level = -1;

    public boolean isCousins(TreeNode root, int x, int y) {
        dfs(root, x, y, 0);
        System.out.println(shared_parent);
        if (x_level > 0 && y_level > 0 && x_level == y_level && !shared_parent) {
            return true;
        }

        return false;
    }

    private void dfs(TreeNode root, int x, int y, int level) {
        if (root == null) return;

        // early stop
        if (shared_parent) return;
        if (x_level > 0 && y_level > 0 && x_level != y_level) return

        if (root.val == x) {
            x_level = level;
            return;
        }
        if (root.val == y) {
            y_level = level;
            return;
        }

        if  (root.left != null && root.right != null) {
            System.out.println(root.left.val + ", " + root.right.val);
            if ( (root.left.val == x && root.right.val == y) ||
                 (root.left.val == y && root.right.val == x)
               ) {
                shared_parent = true;
                return;
            }
        }

        dfs(root.left, x, y, level+1);
        dfs(root.right, x, y, level+1);
    }
}