LeetCode112 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. （Easy）

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.

分析：

还是用递归的思路，可以把sum - root -> val用在递归函数中，这样写最简洁。

代码：

 class Solution {
 public:
     bool hasPathSum(TreeNode* root, int sum) {
         if(root == nullptr) {
             return false;
         }
         if(root -> right == nullptr && root -> left == nullptr) {
             return sum == root->val;
         }
         return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val) ;
     }
 };