112. Path Sum

简单DFS

/** * 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:    vector<int> probVal(TreeNode* root)    {        vector<int> result;        if(root==NULL)            return result;        if(root->left==NULL&&root->right==NULL)        {            result.push_back(root->val);            return result;        }        vector<int> leftResult=probVal(root->left);        vector<int> rightResult=probVal(root->right);        for(int i=0;i<leftResult.size();i++)            result.push_back(root->val+leftResult[i]);        for(int i=0;i<rightResult.size();i++)            result.push_back(root->val+rightResult[i]);        return result;    }    bool hasPathSum(TreeNode* root, int sum) {        vector<int> result=probVal(root);        for(int i=0;i<result.size();i++)        {            if(result[i]==sum)                return true;        }        return false;    }};