113-Path Sum II

类别：DFS

难度：medium

题目描述

算法分析

与112的算法基本一样，但是要注意在实现的过程中对于左子树和右子树path分别进行复制，因为if是先后进行的，第一个if判断后会在第二个if中使用path，而此时path已经发生改变

代码实现

class Solution {
public:
void findPath(vector<vector<int>>& result, vector<int> path, TreeNode* root, int& sum) {
if (root->left == NULL && root->right == NULL) {
int n = path.size();
int count = 0;
for (int i = 0; i < n; ++i) {
count += path[i];
}
if (count == sum) {
result.push_back(path);
}
}
//  notice here copy the path, because after judge root->left, it still use path to judge root->right
vector<int> leftPath = path;
vector<int> rightPath = path;
if (root->left != NULL) {
leftPath.push_back(root->left->val);
findPath(result, leftPath, root->left, sum);
}
if (root->right != NULL) {
rightPath.push_back(root->right->val);
findPath(result, rightPath, root->right, sum);
}
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> result;
if (root == NULL) return result;
vector<int> path;
path.push_back(root->val);
findPath(result, path, root, sum);
}
};