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：

1 class Solution {
2 public:
3     bool hasPathSum(TreeNode* root, int sum) {
4         if(root == nullptr) {
5             return false;
6         }
7         if(root -> right == nullptr && root -> left == nullptr) {
8             return sum == root->val;
9         }
10         return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val) ;
11     }
12 };