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]
]
题目的意思就是对二叉树进行层序遍历,C#的思路是用一个两层的List,每一个用一个队列缓存数据。
public class Solution
{
public IList<IList<int>> LevelOrder(TreeNode root)
{
if(root == null) return null;
// IList<IList<int>> result = new List<List<int>>();
IList<IList<int>> result = new List<IList<int>>();
Queue<TreeNode> tmpQueue = new Queue<TreeNode>();
tmpQueue.Enqueue(root);
while(tmpQueue.Count != 0)
{
List<int> tmpEle = new List<int>();
//
int size = tmpQueue.Count;
for (int i=0 ;i<size; i++) //这里千万不要写 i<tmpQueue.Count 要把tmpQueue 的长度单独拿出来,因为循环完毕之后长度变了
{
TreeNode tmpNode = tmpQueue.Peek();
tmpEle.Add(tmpNode.val);
tmpQueue.Dequeue();
if(tmpNode.left != null) tmpQueue.Enqueue(tmpNode.left);
if (tmpNode.right != null) tmpQueue.Enqueue(tmpNode.right);
}
result.Add(tmpEle);
}
return result;
}
}