Question
Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).
Example 1:
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| Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
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Example 2:
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| Input: root = [1]
Output: [[1]]
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Example 3:
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| Input: root = []
Output: []
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Solution
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| class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
if(root != null) q.offer(root);
while(!q.isEmpty()){
int size = q.size();
List<Integer> level = new ArrayList<>();
for(int i = 0; i < size; i++){
TreeNode cur = q.poll();
level.add(cur.val);
if(cur.left != null) q.offer(cur.left);
if(cur.right != null) q.offer(cur.right);
}
res.add(level);
}
return res;
}
}
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Complexity:
Time complexity: O(n)
Space complexity: O(n)