LeetCode 102. Binary Tree Level Order Traversal

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],

3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

本题目应该使用宽度有限搜索：DFS的方法来处理更为简单

/** * 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>> list = new ArrayList<>();        if (root == null) {            return list;        }        Queue<TreeNode> queue = new LinkedList<>();        queue.offer(root);        while(!queue.isEmpty()) {            int size = queue.size();            List<Integer> level = new ArrayList<>();            for (int i = 0; i < size; i++) {                TreeNode node = queue.poll();                level.add(node.val);                if (node.left != null) {                    queue.offer(node.left);                }                if (node.right != null) {                    queue.offer(node.right);                }            }            list.add(level);        }        return list;    }}