C++递归6行代码
二叉树中和为某一值的路径(一)
http://www.nowcoder.com/practice/508378c0823c423baa723ce448cbfd0c
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
class Solution {
public:
/**
*
* @param root TreeNode类
* @param sum int整型
* @return bool布尔型
*/
bool hasPathSum(TreeNode* root, int sum) {
if(!root) return false;
sum -= root->val;
if(!root->left && !root->right){
if(sum == 0) return true;
else return false;
}
return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
}
};