【leetcode】112. Path Sum

一、题目描述

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.

题目解读：给出一个数sum，从树种寻找是否存在一个从根节点都叶子节点的值的和等于sum的路径

思路：递归的方法很容易解决

c++代码（12ms，22.49%）

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
int total = 0;
if(root == NULL) //注意输入为[] 0的情况，是要返回false的
return false;
else if(root->left==NULL && root->right==NULL)
return root->val == sum;
else{
if(fabs(total) > fabs(sum)) //这步好像多余了，不过写上显得思路更清晰
//当值的大小已经大于sum时，是没有必要再继续算下去的，加上绝对值是应对负数的情况
return false;
else{
total+=root->val;
return hasPathSum(root->left,sum-total) || hasPathSum(root->right, sum-total);
}
}
}
};