https://github.com/emahtab/binary-tree-zigzag-level-order-traversal
Binary Tree Zigzag (Spiral) Order Traversal
https://github.com/emahtab/binary-tree-zigzag-level-order-traversal
bfs binary-tree leetcode problem-solving
Last synced: 3 months ago
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Binary Tree Zigzag (Spiral) Order Traversal
- Host: GitHub
- URL: https://github.com/emahtab/binary-tree-zigzag-level-order-traversal
- Owner: eMahtab
- Created: 2020-02-05T18:20:49.000Z (over 5 years ago)
- Default Branch: master
- Last Pushed: 2020-02-05T18:23:53.000Z (over 5 years ago)
- Last Synced: 2025-02-02T03:25:50.048Z (5 months ago)
- Topics: bfs, binary-tree, leetcode, problem-solving
- Homepage:
- Size: 3.91 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Binary Tree Zigzag Level Order Traversal
## https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
```
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
```
## Implementation :
```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List> zigzagLevelOrder(TreeNode root) {
List> list = new ArrayList<>();
if(root == null)
return list;
Queue queue = new LinkedList<>();
queue.offer(root);
int currentLevel = 0;
while(!queue.isEmpty()){
currentLevel++;
List level = new ArrayList<>();
int size = queue.size();
for(int i = 0; i < size; i++){
TreeNode current = queue.poll();
level.add(current.val);
if(current.left != null)
queue.offer(current.left);
if(current.right != null)
queue.offer(current.right);
}
if(currentLevel % 2 == 0)
Collections.reverse(level);
list.add(level);
}
return list;
}
}
```
# References :
https://www.youtube.com/watch?v=smjr2ow6oKc (Alternate approach using two stacks)