leetcode 113. 路径总和 II(Path Sum II)

目录

• 题目描述：
• 示例:
• 解法：

题目描述：

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

给定如下二叉树，以及目标和 sum = 22，

5/ \4   8/   / \11  13  4/  \    / \7    2  5   1

返回:

[[5,4,11,2],[5,8,4,5]
]

---

解法：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:void pathSum(TreeNode* root, int sum, int cur, vector<vector<int>>& res, vector<int>& path){if(root == NULL){return;}else{path.push_back(root->val);cur += root->val;if(cur == sum && root->left == NULL && root->right == NULL){res.push_back(path);}if(root->left){pathSum(root->left, sum, cur, res, path);}if(root->right){pathSum(root->right, sum, cur, res, path);}path.pop_back();}}vector<vector<int>> pathSum(TreeNode* root, int sum) {vector<vector<int>> res;vector<int> path;int cur = 0;pathSum(root, sum, cur, res, path);return res;}
};