static void Main(string[] args)
{
/*Latest update: July 23, 2015
* Test the code
*/
TreeNode n1 = new TreeNode(1);
n1.left = new TreeNode(2);
n1.right = new TreeNode(3);
n1.left.left = new TreeNode(4);
n1.left.right = new TreeNode(5);
n1.right.left = new TreeNode(6);
n1.right.right = new TreeNode(7);
IList<IList<int>> result3 = levelOrder(n1);
}
/*
* source code from the blog:
* http://bangbingsyb.blogspot.ca/2014/11/leetcode-binary-tree-level-order.html
* comment from blog:
* 这两题实际是一回事，考察的都是树的层序访问。无非就是在I得到结果后做一次reverse就能得到
* II要求的结果。无论是树还是图，层序访问最直接的方法就是BFS。
1. BFS可以用一个queue实现，但是难以跟踪每个节点究竟是在第几层。解决办法是除了压入节点
指针外，还同时压入节点所在的层数，即压入pair<TreeNode*, int>。
* Julia's comment:
* 1. convert C++ code to C# code
* 2. 单个queue的迭代解法 - using queue, iteratively solution
* 3. 在压入root后，需要额外压入一个NULL来标记第i层尾, 最后一层不用.
*/
public static IList<IList<int>> levelOrder(TreeNode root)
{
IList<IList<int>> levelNodeValues = new List<IList<int>>();
if (root == null) return levelNodeValues;
Queue<TreeNode> q = new Queue<TreeNode>();
q.Enqueue(root);
q.Enqueue(null);
int curLevel = 0;
while (q.Count >0)
{
TreeNode cur = q.Peek();
q.Dequeue();
if (cur == null)
{
curLevel++;
if (q.Count > 0) // ensure the checking, avoid dead loop
q.Enqueue(null);
}
else
{
if (curLevel == levelNodeValues.Count)
levelNodeValues.Add(new List<int>());
levelNodeValues[curLevel].Add(cur.val);
if (cur.left != null)
q.Enqueue(cur.left);
if (cur.right != null)
q.Enqueue(cur.right);
}
}
return levelNodeValues;
}
public TreeNode(int x)
{
val = x;
left = null;
right = null;
}
