main.cpp

// Source: https://leetcode.com/problems/root-equals-sum-of-children
// Title: Root Equals Sum of Children
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// You are given the `root` of a **binary tree** that consists of exactly `3` nodes: the root, its left child, and its right child.
//
// Return `true` if the value of the root is equal to the **sum** of the values of its two children, or `false` otherwise.
//
// **Example 1:**
// https://assets.leetcode.com/uploads/2022/04/08/graph3drawio.png
//
// ```
// Input: root = [10,4,6]
// Output: true
// Explanation: The values of the root, its left child, and its right child are 10, 4, and 6, respectively.
// 10 is equal to 4 + 6, so we return true.
// ```
//
// **Example 2:**
// https://assets.leetcode.com/uploads/2022/04/08/graph3drawio-1.png
//
// ```
// Input: root = [5,3,1]
// Output: false
// Explanation: The values of the root, its left child, and its right child are 5, 3, and 1, respectively.
// 5 is not equal to 3 + 1, so we return false.
// ```
//
// **Constraints:**
//
// - The tree consists only of the root, its left child, and its right child.
// - `-100 <= Node.val <= 100`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

using namespace std;

struct TreeNode {
  int val;
  TreeNode *left;
  TreeNode *right;
  TreeNode() : val(0), left(nullptr), right(nullptr) {}
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
  TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
 public:
  bool checkTree(TreeNode *root) {  //
    return root->val == root->left->val + root->right->val;
  }
};