【leetcode】Path Sum II
Question：
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 1return
[
[5,4,11,2],
[5,8,4,5]
]Anwser 1：
/**
* 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 calPath(TreeNode *root, int sum, vector<int> tmp, vector< vector<int> > &ret){
if(root == NULL){
return;
}
tmp.push_back(root->val);
if(sum == root->val && root->left == NULL && root->right == NULL){
ret.push_back(tmp);
}
calPath(root->left, sum - root->val, tmp, ret);
calPath(root->right, sum - root->val, tmp, ret);
tmp.pop_back();
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector< vector<int> > ret;
if(root == NULL) {
return ret;
}
vector<int> tmp;
calPath(root, sum, tmp, ret);
return ret;
}
};Anwser 2：
/**
* 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 calPath(TreeNode *root, int sum, vector<int> tmp, vector< vector<int> > &ret){
if(root == NULL){
return;
}
tmp.push_back(root->val);
if(sum == root->val && root->left == NULL && root->right == NULL){
ret.push_back(tmp);
tmp.pop_back(); // pop tmp vector
return;
}
calPath(root->left, sum - root->val, tmp, ret);
calPath(root->right, sum - root->val, tmp, ret);
tmp.pop_back();
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector< vector<int> > ret;
if(root == NULL) {
return ret;
}
vector<int> tmp;
calPath(root, sum, tmp, ret);
return ret;
}
};本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2013年04月13日，如有侵权请联系 cloudcommunity@tencent.com 删除
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