LeetCode:Path Sum II

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]
]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,sum - rt->val,ans,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,ans,path);return ans ;}
};

转载于:https://www.cnblogs.com/ldxsuanfa/p/10515912.html

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