257. Binary Tree Paths

Posted on Edited on In

257. Binary Tree Paths

Description

Given the root of a binary tree, return all root-to-leaf paths in any order .

A leaf is a node with no children.

Example 1:

1
2
Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]

Example 2:

1
2
Input: root = [1]
Output: ["1"]

Constraints:

  • The number of nodes in the tree is in the range [1, 100].
  • -100 <= Node.val <= 100

Hints/Notes

  • traverse the tree

Solution

Language: C++

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
36
37
38
39
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<string> res;

    vector<string> binaryTreePaths(TreeNode* root) {
        traverse(root, "");
        return res;
    }

    void traverse(TreeNode* root, string path) {
        if (!root) {
            return;
        }
        path += to_string(root->val);
        bool node = true;
        if (root->left) {
            traverse(root->left, path + "->");
            node = false;
        }
        if (root->right) {
            traverse(root->right, path + "->");
            node = false;
        }
        if (node) {
            res.push_back(path);
        }
    }
};