leetcode 113. 路径总和 II（C++、python）

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:
给定如下二叉树，以及目标和 sum = 22，

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

返回:

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

C++

/**
 * 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, int sum, int val, vector<int>& tmp, vector<vector<int>>& res)
    {
        if(NULL==root)
        {
            return;
        }
        else
        {
            val+=root->val;
            tmp.push_back(root->val);
            if(NULL==root->left && NULL==root->right)
            {
                if(sum==val)
                {
                    res.push_back(tmp);
                }
            }
            else
            {
                dfs(root->left,sum,val,tmp,res);
                dfs(root->right,sum,val,tmp,res);
            }
            tmp.pop_back();
            val-=root->val;
            
        }
    }
    
    vector<vector<int>> pathSum(TreeNode* root, int sum) 
    {
        vector<vector<int>> res;
        vector<int> tmp;
        if(NULL==root)
        {
            return res;
        }
        else
        {
            dfs(root,sum,0,tmp,res);
            return res;
        }                
    }
};

python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def dfs(self,root,val,su,res,tmp):
        if root:
            if None==root.left and None==root.right:
                if su==val+root.val:
                    res.append(tmp+[root.val])
            else:
                self.dfs(root.left,val+root.val,su,res,tmp+[root.val])
                self.dfs(root.right,val+root.val,su,res,tmp+[root.val])
            
    def pathSum(self, root: TreeNode, su: int) -> List[List[int]]:
        res=[]
        tmp=[]
        if None==root:
            return res
        else:
            self.dfs(root,0,su,res,tmp)
            return res