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         //其实就是比较节点值和 不断减小的sum值 是否相同
14         //1、空节点
15         if(root==NULL) return false;
16         //2、单节点
17         if(root->left == NULL && root->right==NULL)
18             return root->val == sum;
19         //3、左节点或者右节点为空，那就递归右子树或者左子树
20         if(root->left==NULL) return hasPathSum(root->right,sum - root->val);
21         if(root->right==NULL) return hasPathSum(root->left,sum- root->val);
22         //左右子树都不为空
23         return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val);
24         
25     }
26 };