0102. Binary Tree Level Order Traversal - kumaeki/LeetCode GitHub Wiki

0102. Binary Tree Level Order Traversal

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:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -1000 <= Node.val <= 1000

解法1

题目本身很简单. 就是按照层遍历tree.

/**
 * 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>> res = new ArrayList<List<Integer>>();
        if(root == null)
            return res;
        ArrayDeque<TreeNode> que = new ArrayDeque<TreeNode>();
        que.offer(root);
        while(!que.isEmpty()){
            int s = que.size();
            List<Integer> list = new ArrayList<Integer>();
            while(s > 0){
                TreeNode node = que.poll();
                list.add(node.val);
                if(node.left != null)
                    que.offer(node.left);
                if(node.right != null)
                    que.offer(node.right);
                s--;
            }
            res.add(list);
        }
        return res;
    }
}
⚠️ **GitHub.com Fallback** ⚠️