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]
]
分析:
这道题目算是Path Sum的延伸。仅仅只是要求出全部从根到树叶的路径上节点和为给定值的路径,并放到一个vector中返回。
採取相似于Path Sum递归思路。每次递归都会建立一个新的vector,复制之前的vector内容,并将当前节点的值放入。假设当前节点是叶节点。而且路径上节点和满足要求。就将新的vector放入终于返回的vector中。
代码:
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> pathSum(TreeNode* root, int sum) {
if(!root)
return res;
vector<int> tmp;
pathSum1(root,sum,tmp);
return res;
}
void pathSum1(TreeNode* root, int sum,vector<int> tmp)
{
if(!root)
return ;
vector<int> tmp1;
tmp1 = tmp;
tmp1.push_back(root->val);
if(!root->left&&!root->right)
{
if(root->val==sum)
res.push_back(tmp1);
return;
}
pathSum1(root->left,sum-root->val,tmp1);pathSum1(root->right,sum-root->val,tmp1);
}
};