/*
Latest update: July 23, 2015
* Leetcode question: Binary tree level order traversal
*
* 1. Practice more using Interface instead of actual class, like ArrayList in function return
* 2. Change the return using IList<IList<int>>
* 3. Test the code using leetcode online judge
* Read the website about ArrayList vs List:
* https://social.msdn.microsoft.com/Forums/vstudio/en-US/37c64e16-69e9-4935-abaa-ee8987230240/arraylist-vs-ilistobject?forum=netfxbcl&prof=required
* julia now understands the importance to use the interface, more generic, and less type casting,
* quick running time.
*
* Leetcode online result:
* 34 / 34 test cases passed.
Status: Accepted
Runtime: 536 ms
*/
public static IList<IList<int>> levelOrder(TreeNode root)
{
IList<IList<int>> result = new List<IList<int>>();
if (root == null)
return (IList<IList<int>>)result;
List<int> level = new List<int>();
Queue<TreeNode> toVisit = new Queue<TreeNode>();
toVisit.Enqueue(root);
toVisit.Enqueue(null);
while (true)
{
TreeNode cur = toVisit.Peek();
toVisit.Dequeue();
// end a level
if (cur == null)
{
List<int> list = new List<int>(); // julia: bug fix
list.AddRange(level); // julia: bug fix
result.Add(list);
if (toVisit.Count == 0)
break;
// add a new End-Of-Level
level.Clear();
toVisit.Enqueue(null);
}
else
{
level.Add(cur.val);
if (cur.left != null)
toVisit.Enqueue(cur.left);
if (cur.right != null)
toVisit.Enqueue(cur.right);
}
}
return (IList<IList<int>>)result;
}
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);
}
static void Main(string[] args)
{
/*Latest update: July 23, 2015
* Test the code and then find a bug
*/
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);
ArrayList result = levelOrder_bugVersion(n1);
/*
* Try to fix the bug
*
*/
ArrayList result2 = levelOrder_bugFree(n1);
}
public TreeNode(int x)
{
val = x;
left = null;
right = null;
}
/**
* source code from blog:
* https://github.com/xiaoxq/leetcode-cpp/blob/master/src/BinaryTreeLevelOrderTraversal.cpp
* convert from C++ to C#
* julia's comment:
* 1. check leetcode online judge, need to use IList<IList<int>>
* 2. how to use queue for level order traversal, null point as level divider. Kind of tricky.
* first action before the loop, push first node into the queue, and then
* push level divider: null point
* design null pointer action: once null pointer is met, and then,
* finish the current level, go to next level:
* 1. first, save the current level visited value into the result for return;
* 2. second, check if the queue is empty -> yes, finish last level, ready to exit.
* 3. third, prepare for next level:
* A: clear the container for current level info
* B: review facts: B.1. there is no null in the queue
* B.2. there is only one level nodes in the queue, will be current level
* So, need to add level divider now.
* 3. C# has a bug - share one arrayList object for each level,
* use clear method will not work out.
* lesson A: C# List.addAll <-> C# ArrayList.addRange
*/
public static ArrayList levelOrder_bugVersion(TreeNode root)
{
ArrayList result = new ArrayList();
if( root == null)
return result;
ArrayList level = new ArrayList();
Queue<TreeNode> toVisit = new Queue<TreeNode>();
toVisit.Enqueue( root );
toVisit.Enqueue( null );
while( true )
{
TreeNode cur = toVisit.Peek();
toVisit.Dequeue();
// end a level
if( cur==null )
{
result.Add( level );
if( toVisit.Count == 0 )
break;
// add a new End-Of-Level
level.Clear(); // julia's comment: C# bug, cannot share one object -
//need to create new object for each level
toVisit.Enqueue( null );
}
else
{
level.Add( cur.val );
if( cur.left != null)
toVisit.Enqueue( cur.left );
if( cur.right != null )
toVisit.Enqueue( cur.right );
}
}
return result;
}