leetcode37:path-sum-ii

题目描述

给定一个二叉树和一个值sum，请找出所有的根节点到叶子节点的节点值之和等于sum的路径，
例如：
给出如下的二叉树，sum=22，
    5
    / \
  4  8
  /    / \
11  13 4
/ \      / \

7 2   5 1       
返回
[
    [5,4,11,2],
    [5,8,4,5]
]

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 andsum = 22,

    5
    / \
  4  8
  /    / \
11  13 4
/ \      / \

7 2   5 1    

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

示例1

输入

复制

{1,2},1

输出

复制

[]

示例2

输入

复制

{1,2},3

输出

复制

[[1,2]]

/**
 * struct TreeNode {
 *    int val;
 *    struct TreeNode *left;
 *    struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > pathSum(TreeNode* root, int sum) {
        // write code here
        vector <vector <int>> vv;
        vector <int> v;
        pathSum_Aux(root,sum,v,vv);
        return vv;
    }
    void pathSum_Aux(TreeNode *root,int sum,vector <int> v,vector<vector<int>>&vv){
        if (root==NULL)
            return ;
        v.push_back(root->val);
        if (root->left ==NULL && root->right==NULL &&sum -root->val==0){
            vv.push_back(v);
        }
        pathSum_Aux(root->left, sum-root->val, v,vv);
        pathSum_Aux(root->right,sum-root->val,v,vv);
        
    }
};
/**
 * struct TreeNode {
 *    int val;
 *    struct TreeNode *left;
 *    struct TreeNode *right;
 * };
 */

class Solution {
    vector <vector<int>>res;
    void dfspath(TreeNode *root,int num,vector<int>v){
        if (!root)return ;
        v.push_back(root->val);
        if (root->left ==nullptr && root->right==nullptr){
            if (num==root->val )res.push_back(v);
            
        }
        dfspath(root->left,num-root->val,v);
        dfspath(root->right,num-root->val,v);
    }
public:
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > pathSum(TreeNode* root, int sum) {
        // write code here
        vector <int>v;
        dfspath(root, sum, v);
        return res;
        
    }
};
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

#
# 
# @param root TreeNode类 
# @param sum int整型 
# @return int整型二维数组
#
class Solution:
    def pathSum(self , root , sum ):
        # write code here
        if not root:
            return []
        res=[]
        
        def helper(root,remain,temp=[]):
            temp_=temp+[root.val]
            remain_=remain-root.val
            if remain_==0 and not root.left and not root.right:
                res.append(temp_)
                return 
            if root.left:
                helper(root.left,remain_,temp_)
            if root.right:
                helper(root.right,remain_,temp_)
        helper(root,sum)
        return res