LeetCode Path Sum 判断树的路径之和

 1 /**
 2  * Definition for binary tree
 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==0)        //针对根结点，非根节点不应会执行到这里
14          return false;
15 
16      if( root->left==0 && root->right==0 )        //是叶子结点，且刚好将sum变为0
17          return root->val==sum ? true : false ;
18 
19      if( ( root->left!=0 && hasPathSum( root->left , sum-root->val) ) || ( root->right!=0 && hasPathSum( root->right , sum-root->val ) ) )//判断左叉或右叉中是否有一条从上往下满足要求的路径
20         return true;
21 
22      return false;
23     }
24 };