LeetCode 113. 路徑總和 II 二叉樹搜索+記錄路徑

思路：

當搜索到葉子節點時，我們看當前的s是否爲0即可。

代碼：

/**
 * 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:
    void dfs(TreeNode* root,vector<vector<int>> &ans,vector<int>path,int sum)
    {
        if(!root)return ;
        path.push_back(root->val);
        int x=sum-root->val;//先減去這個節點的值，看是否符合條件
        if(!root->left&&!root->right&&!x)
        {
            ans.push_back(path);
        }
        dfs(root->left,ans,path,x);
        dfs(root->right,ans,path,x);
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> ans;
        vector<int>path;
        dfs(root,ans,path,sum);
        return ans;
    }
};