【二叉树】113. Path Sum II

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=。=