113. Path Sum II (Medium)

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

 

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: []

Example 3:

Input: root = [1,2], targetSum = 0
Output: []

 

Constraints:

• The number of nodes in the tree is in the range [0, 5000].
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

A:       就是简单的递归调用

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        vector<int> tmp;
        vector<vector<int>> res;
        helper(root, targetSum, tmp, res);
        return res;
    }
private:
    void helper(TreeNode* root, int targetSum, vector<int> & tmp, vector<vector<int>> & res){
        if(!root)
            return;
        tmp.push_back(root->val);
        targetSum -= root->val;
        if(!root->left && !root->right){ // is leaf node
            if(targetSum == 0){
                res.push_back(vector<int>(tmp));
            }
        }
        helper(root->right, targetSum, tmp, res);
        helper(root->left, targetSum, tmp, res);
        tmp.pop_back();
    }
};

Mistakes:
1:  当到了leaf节点，并且相等的时候，忘了返回，  艹
2:  ，当不满足的时候，要返回空，而不是null。


------------------------第二遍-----------------
和第一遍不一样的地方就在于， 重新弄了一个函数， 把all 作为参数传过去了。