LeetCode 993. Cousins in Binary Tree（判斷結點是否爲Cousin）

993. Cousins in Binary Tree

In a binary tree, the root node is at depth 0, and children of each depth knode 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 yare 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

思路：

求解x，y的深度和父親結點，若是深度同樣，父親結點不一樣，就是true；不然，就是false。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *getDepth(TreeNode* root, int val, int depth, int &level)
    {
        if (!root) return nullptr;
        
        //找到val的父親，並設置val的depth
        if ((root->left && root->left->val == val) || (root->right && root->right->val == val))
        {
            level = depth;
            return root;
        }

        TreeNode *left = getDepth(root->left, val, depth + 1, level);
        if (left) return left;

        TreeNode *right = getDepth(root->right, val, depth + 1, level);
        if (right) return right;

        return nullptr;
    }

    bool isCousins(TreeNode *root, int x, int y)
    {
        int x_depth = -1, y_depth = -1;
        TreeNode *x_parent = getDepth(root, x, 0, x_depth);
        TreeNode *y_parent = getDepth(root, y, 0, y_depth);
        if (x_depth == y_depth  && x_parent != y_parent)
        {
            return true;
        }
        return false;
    }
};