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Leetcode dfs Path SumII
Path Sum II
Total Accepted: 18489 Total Submissions: 68323My SubmissionsGiven 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<vector<int> > pathSum(TreeNode *root, int sum) { vector<int> cur; _pathSum(root, sum,cur); return res; } private: vector<vector<int> > res; void _pathSum(TreeNode *root, int sum, vector<int> &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(); } };
Leetcode dfs Path SumII
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