Leetcode dfs Path SumII - c++编程基础

Path Sum II

Total Accepted: 18489 Total Submissions: 68323My Submissions

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]
]

题意：给定一棵二叉树和一个值，在二叉树中找到从根到叶子的路径使得路径中的节点的总值
等于给定值
思路：dfs
复杂度：时间O(n) 空间O(log n)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector
  
    > pathSum(TreeNode *root, int sum) {
    	vector
   
     cur; _pathSum(root, sum,cur); return res; } private: vector
    
      > res; void _pathSum(TreeNode *root, int sum, vector
     
       &path){ if(!root) return ; path.push_back(root->val); if(!root->left && !root->right){ if(root->val == sum) { res.push_back(path); } } _pathSum(root->left, sum - root->val, path); _pathSum(root->right, sum - root->val, path); path.pop_back(); } };