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;
}
}