102. Binary Tree Level Order Traversal #

题目 #

给定二叉树的根节点 root，返回其节点值的 层序遍历。

即逐层地，从左到右访问所有节点。

思路 #

代码 #

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>> ans = new LinkedList<>();
        if (root == null) return ans;
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (queue.size() != 0) {
            ans.add(new LinkedList<>());
            
            for (int i = 0, layerSize = queue.size(); i < layerSize; i++) {
                TreeNode node = queue.poll();
                
                ans.get(ans.size() - 1).add(node.val);
                
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        
        return ans;
    }
}