Leet code Question no: 102
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Oracle 7
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Adobe 2
Docusign 2
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TikTok 2
Intuition
We aim to perform a level-order traversal of a binary tree, collecting nodes at each level into sublists.
Approach
- Utilize a queue for level-order traversal, initializing an empty list
WrapListto store results. - While the queue is not empty, dequeue nodes, collect their values into sublists, and enqueue their children.
- Add sublists to
WrapListas levels are processed and return it as the result.
Complexity
- Time complexity: O(n) - We visit each node once during the traversal.
- Space complexity: O(n) - The space usage grows with the number of nodes due to the queue and the result list.
Code
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> WrapList = new LinkedList<>();
if(root == null)
return WrapList;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
int levelNum = queue.size();
List<Integer> subList = new LinkedList<>();
for(int i = 0 ; i < levelNum ; i++){
TreeNode node = queue.poll();
subList.add(node.val);
if(node.left != null) queue.offer(node.left);
if(node.right != null) queue.offer(node.right);
}
WrapList.add(subList);
}
return WrapList;
}
}
Happy coding,
shiva
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