112. Path Sum

果然还是递归比较好理解，首先判断是否有根节点、根节点的值是否正好等于sum；然后递归调用hasPathSum，注意第二个参数改为sum-root->val:

/** * 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 == NULL) return false;        if (root->val == sum && !(root->left) && !(root->right)) return true;        return (hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val));    }};