Description
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
tree
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
Solution
Level-traversal, time O(n), space O(n)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> levels = new LinkedList<>();
if (root == null) {
return levels;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int levelSize = queue.size();
List<Integer> level = new LinkedList<>();
while (levelSize-- > 0) {
TreeNode node = queue.poll();
level.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
levels.add(level);
}
return levels;
}
}
