Problem Description:
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:
利用递归,一旦到叶子节点且和为sum,记录结果,注意元素进栈后要出栈。
代码如下:
/**
* 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 findpath(TreeNode *root,int sum,vector<int> &nodes,vector<vector<int> > &res)
{
if(!root)
return;
nodes.push_back(root->val);
if(!root->left&&!root->right&&sum==root->val)
{
res.push_back(nodes);
nodes.pop_back();
return;
}
findpath(root->left,sum-root->val,nodes,res);
findpath(root->right,sum-root->val,nodes,res);
nodes.pop_back();
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int> > res;
vector<int> nodes;
findpath(root,sum,nodes,res);
return res;
}
};