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.
判断是否存在一条从根节点到叶子节点的路径，路径上节点之和为sum
C++（6ms）：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool hasPathSum(TreeNode* root, int sum) { 13 if (root == NULL) return false ; 14 if (root->val == sum && root->left == NULL && root->right == NULL) return true ; 15 return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val) ; 16 } 17 };
