LeetCode 112. Path Sum

题目链接：https://leetcode.com/problems/path-sum/?tab=Description

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.

思路一：

最直观的思路就是用DFS遍历，遍历过程中将路径上各节点的值加起来，看最后到叶子节点时，是否等于sum,代码不够简洁

代码：

/** * 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) {        bool flag=0;        int tmp=0;        DFS(root,sum,tmp,flag);        return flag;    }    void DFS(TreeNode* root,int sum,int tmp,bool& flag)    {        if(root==NULL)            return;        tmp+=root->val;        if(tmp==sum&&root->left==NULL&&root->right==NULL)        {            flag=1;            return;        }        DFS(root->left,sum,tmp,flag);        DFS(root->right,sum,tmp,flag);    }};

思路二：

不用另外写一个函数，就用给出的函数进行递归调用，用sum逐步减去遍历到的节点，无论左子树还是右子树满足条件都行，代码很简洁

代码：

/** * 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) {        if(root==NULL)            return 0;        if(sum-root->val==0&&root->left==NULL&&root->right==NULL)            return 1;        return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);    }};