[leetcode] 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.
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]
]

解法一：

/**
* 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<int>> pathSum(TreeNode* root, int sum) {
if(!root) return {};
vector<vector<int>> res;
vector<int> out;
pathSum(res, out, root, sum);
return res;
}

void pathSum(vector<vector<int>>&res, vector<int>& out, TreeNode* root, int sum){
if(!root) return;

if(!root->left&&!root->right){
if(root->val==sum){
out.push_back(root->val);
res.push_back(out);
out.pop_back();
return;
}else return;
}

out.push_back(root->val);
pathSum(res, out, root->left, sum-root->val);
out.pop_back();

out.push_back(root->val);
pathSum(res, out, root->right, sum-root->val);
out.pop_back();
}
};

解法二：

更加简洁点。

/**
* 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<int>> pathSum(TreeNode* root, int sum) {
if(!root) return {};
vector<vector<int>> res;
vector<int> out;
out.push_back(root->val);
pathSum(res, out, root, sum);
return res;
}

void pathSum(vector<vector<int>>&res, vector<int>& out, TreeNode* root, int sum){
if(!root->left&&!root->right&&root->val==sum){
res.push_back(out);
}

if(root->left){
out.push_back(root->left->val);
pathSum(res, out, root->left, sum-root->val);
out.pop_back();
}

if(root->right){
out.push_back(root->right->val);
pathSum(res, out, root->right, sum-root->val);
out.pop_back();
}
}
};