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:
这题并不难。其实就是深搜。
就是可能一开始,觉得函数递归的时候,要传哪些参数下去?要返回那些值??
当有多个返回值的时候怎么办??
Trick: C++ 中可以采用引用,来返回值的功能。
1. 返回值: 多条 paths, 可以将 vector<vector<int> >& result 作为形参; 这样其实 所有的递归函数都共享了这个 vector; 当找到路径时,push_back即可。
2. 需要往下传的参数: sum, 当前的路径;
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int> > result;
vector<int> path; MyPathSum(root, sum, path, result); return result;
}
private:
void MyPathSum(TreeNode *root, int sum, vector<int> &path, vector<vector<int> > &result)
{
if(root == NULL) return; path.push_back(root->val);
if(root->left == NULL && root->right == NULL){
if(sum == root->val){
result.push_back(path);
}
} if(root->left){
MyPathSum(root->left, sum - root->val, path, result);
} if(root->right){
MyPathSum(root->right, sum - root->val, path, result);
}
path.pop_back();
}
};