[LeetCode] Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / 
            4   8
           /   / 
          11  13  4
         /        
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

判断一棵树由根到叶的某一条路径节点值是否等于给定和。

因为要对买个节点值都要处理所以使用递归，结果要求返回一个bool值，只要判断任意一条路径符合就返回true。

这里用sum减去每个节点值来作为判断条件。如果叶节点值等于sum的值返回true。否则返回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:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == nullptr)
            return false;
        if (root->left == nullptr && root->right == nullptr)
            return root->val == sum;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};
// 9 ms