leetcode-java-102. Binary Tree Level Order Traversal

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } *  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 *//*利用广度搜素来实现层次遍历：1.利用queue，while循环知道queue为null结束2.本层元素加入存储每层level的list中，利用queue的poll方法，且利用size来控制每层存储的结束3.下层元素利用offer加到queue中 */ public class Solution {     public List<List<Integer>> levelOrder(TreeNode root) {         List<List<Integer>> list = new LinkedList<List<Integer>>();         Queue<TreeNode> queue = new LinkedList<TreeNode>();         queue.offer(root);        // 此行需要加，因为不加就算伪null也会进入while循环，因为null加进去了，并且m=1         if(root == null) {             return list;         }         while(!queue.isEmpty()) {             int m = queue.size();             List<Integer> level = new LinkedList<Integer>();            // 将特定层的数据存于list中,下一层存于queue中             while(m > 0) {                 TreeNode node = queue.peek();                 if(node.left != null) {                     queue.offer(node.left);                 }                 if(node.right != null) {                     queue.offer(node.right);                 }                 level.add(queue.poll().val);                 m--;             }             list.add(level);         }         return list;     } }