标签：post integer lin rom size blog ott code form
题目：
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] ]
题意及分析：宽度遍历一棵树。用队列保存中间值即可。
代码:
/**
* 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>> res = new ArrayList<>();
if(root==null) return res;
List<Integer> list = new ArrayList<>();
Queue<TreeNode> queue=new LinkedList<>();
queue.offer(root);
Queue<TreeNode> queue1=new LinkedList<>();
while(!queue.isEmpty()||!queue1.isEmpty()){
queue1.clear();
while(!queue.isEmpty()){ //遍历一层
TreeNode now = queue.poll();
list.add(now.val);
if(now.left!=null)
queue1.offer(now.left);
if(now.right!=null)
queue1.offer(now.right);
}
queue=new LinkedList<>(queue1); //进行下一层的遍历
res.add(new ArrayList<>(list));
list.clear(); //清空
}
return res;
}
}
[LeetCode] 102. Binary Tree Level Order Traversal Java
标签：post integer lin rom size blog ott code form
原文地址：http://www.cnblogs.com/271934Liao/p/7197331.html
