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(); } };