Leetcode solution 257: Binary Tree Paths
Blogger:https://blog.baozitraining.org/2019/08/leetcode-solution-257-binary-tree-paths.html
Youtube: https://youtu.be/1Ge9R_L9dt8
部落格園: https://www.cnblogs.com/baozitraining/p/11441305.html
B站: https://www.bilibili.com/video/av66230678/
Problem Statement
Given a binary tree, return all root-to-leaf paths.
Note: A leaf is a node with no children.
Example:
Input:
1
/ \
2 3
\
5
Output: ["1->2->5", "1->3"]
Explanation: All root-to-leaf paths are: 1->2->5, 1->3
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Problem link
Video Tutorial
You can find the detailed video tutorial here
- Youtube
- B站
Thought Process
Easy problem. Use pre-order traversal from root to leaf and keep the recursion path. The recursion returning condition would be when a node doesn’t have left and right children. Note use a string to keep appending is easier than using a string builder, because we need to pop out and reset the string builder.
Caveat
- Handle the single node situation when no "->". E.g., A vs A->B
There is also an iterative implementation of this using one stack, similar to BST iterator using one stack problem.
Solutions
Pre-order traversal
1 private List<String> res; // stores the final output
2
3 public List<String> binaryTreePaths(TreeNode root) {
4 this.res = new ArrayList<>();
5 helper(root, "");
6 return this.res;
7 }
8
9 // helper function that does basic depth first traversal
10 private void helper(TreeNode root, String str) {
11 if(root == null) {
12 return;
13 }
14
15 if(root.left==null && root.right==null) // reach a leaf node, thus completes a path and need to add it into the output
16 this.res.add(str + root.val);
17 else {
18 str += root.val + "->";
19 helper(root.left, str);
20 helper(root.right, str);
21 }
22 return;
23 }
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Time Complexity: O(N), each node is visited once
Space Complexity: O(N), N is the total nodes since that's the string length you need
References
- Leetcode official solution