【leetcode】【medium】113. Path Sum II
113. 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.
Note: A leaf is a node with no children.
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]
]
题目链接：https://leetcode-cn.com/problems/path-sum-ii/
思路
法一：递归
将treePath和pathSum两道题的思路综合，能理解前两题的思路，就很容易得到本题思路。
/**
* 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:
vector<vector> pathSum(TreeNode* root, int sum) {
vector<vector> res;
if(!root) return res;
if(!root->left && !root->right){
if (sum == root->val){
vector tmp = {root->val};
res.push_back(tmp);
}
return res;
}
vector<vector> l = pathSum(root->left, sum-root->val);
for(auto item: l){
item.insert(item.begin(), root->val);
res.push_back(item);
}
vector<vector> r = pathSum(root->right, sum-root->val);
for(auto item: r){
item.insert(item.begin(), root->val);
res.push_back(item);
}
return res;
}
};
法二：非递归
非递归的实现方法有很多：
1）用queue/stack实现的前/中/后/层序遍历。
2）用前驱后继节点实现的DFS。
这里为了练习递归思想，暂时不实现非递归方法了，后序可能会再补上。
作者：lemonade13
