LeetCode之102. Binary Tree Level Order Traversal

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;
        }
    }