【二叉树】113. Path Sum II

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5             / \            4   8           /   / \          11  13  4         /  \    / \        7    2  5   1

return

[   [5,4,11,2],   [5,8,4,5]]

和Binary Tree Paths类似的，但是并没有想象中容易，因为顺序搜下来很难记录总和和此时vector中的内容，看了Discuss。。。。重点就在sum-root->val那里，解答：

/** * 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<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>>ans;        vector<int> path;        if(!root)return ans;        dfs(root,sum,ans,path);        return ans;    }    void dfs(TreeNode* root, int sum,vector<vector<int>>& ans,vector<int>& path){        path.push_back(root->val);        if(!(root->left)&&!(root->right)&&(sum==root->val))            ans.push_back(path);        if(root->left)            dfs(root->left,sum-root->val,ans,path);        if(root->right)            dfs(root->right,sum-root->val,ans,path);        path.pop_back();    }};

然后注意传参的时候要传引用，不要粗心=。=

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