LeetCode112 Path Sum

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. (Easy)

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.

分析:

还是用递归的思路,可以把sum - root -> val用在递归函数中,这样写最简洁。

代码:

 1 class Solution {
 2 public:
 3     bool hasPathSum(TreeNode* root, int sum) {
 4         if(root == nullptr) {
 5             return false;
 6         }
 7         if(root -> right == nullptr && root -> left == nullptr) {
 8             return sum == root->val;
 9         }  
10         return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val) ;
11     }
12 };
原文地址:https://www.cnblogs.com/wangxiaobao/p/6044961.html