LeetCode:Path Sum II
发布时间:2020-12-14 04:37:40 所属栏目:大数据 来源:网络整理
导读:????? 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],[5,8,5]]解题思路: 通
????? 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 5
/ 4 8
/ / 11 13 4
/ / 7 2 5 1
return [
[5,4,11,2],[5,8,5]
]
解题思路:
通过遍历树保存从根到叶子节点的路径和路径和。然后推断其和是否等于sum就可以.
解题代码:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x),left(NULL),right(NULL) {} * }; */ class Solution { public: void dfs(TreeNode *rt,long long sum,vector<vector<int> > &ans,vector<int> &path) { path.push_back(rt->val); if (rt->left == rt->right && rt->left == NULL) { if (sum == rt->val) ans.push_back(vector<int>(path.begin(),path.end())); path.erase(path.end()-1); return ; } if (rt->left) dfs(rt->left,sum - rt->val,ans,path); if (rt->right) dfs(rt->right,path); path.erase(path.end()-1); } vector<vector<int> > pathSum(TreeNode *root,int sum) { vector<vector<int> > ans ; if (!root) return ans ; vector<int> path; dfs(root,sum,path); return ans ; } };? (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |