Leetcode--Path Sum I,II,III

I. Question

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:

Given the below binary tree and sum = 22,

5

/ \

4 8

/ / \

11 13 4

/ \ \

7 2 1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution

想法很简单，分别递归左右子树即可，但是要注意返回条件判断。

class Solution {

public:
bool fun(TreeNode* root, int sum){
if (root == NULL)
return false;
//返回条件在前一个结点就判断，防止出现[1,2],1返回true的情况
if (root->left == NULL&&root->right == NULL){
if (sum == root->val)
return true;
else
return false;
}
int temp = sum - root->val;
//返回左右子树的或
return (fun(root->left, temp) || fun(root->right, temp));
}
bool hasPathSum(TreeNode* root, int sum) {
if (root == NULL)
return false;
return fun(root, sum);
}
};

II. Question

在上一题基础上，要求输出所有符合条件的路径，默认路径存在。想到用引用来实现二维路径的存储，关于返回值的问题还需加深理解。

class Solution {
public:
bool fun(TreeNode *root, int sum, vector<vector<int>> &paths, vector<int> path){
if (root == NULL)
//path.pop_back();
return 0;
else if (root->left == NULL&&root->right == NULL){
if (sum == root->val){
path.push_back(root->val);
paths.push_back(path);
//  path.clear();
return 0;
}
}
int temp = sum - root->val;
path.push_back(root->val);
fun(root->left, temp, paths,path);
fun(root->right, temp, paths,path);
return 0;
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> paths;
vector<int> path;
fun(root, sum, paths, path);
return paths;
}
};