Binary Tree Paths

Desicription

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

1
2
3
4
5
6
7
8
9
10
11
Input:

1
/ \
2 3
\
5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
/**
* 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 {
private:
vector<string> res{};
void search(TreeNode* root, string path) {
if(root == nullptr) {
return ;
}
path += path == "" ? to_string(root->val) : "->" + to_string(root->val);
if(root->left == nullptr && root->right == nullptr) {
res.push_back(path);
return ;
}
if(root->left) {
search(root->left, path);
}
if(root->right) {
search(root->right, path);
}
}
public:
vector<string> binaryTreePaths(TreeNode* root) {
search(root, "");
return res;
}
};