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
Given binary tree
[3,9,20,null,null,15,7]
, 3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
/*
BFS 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; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int depth = 0;
while(!queue.isEmpty()) {
depth++;
int size = queue.size();
List<Integer> list = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
list.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
ans.add(list);
}
return ans;
}
}
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