/*
* Leetcode: 100 same tree
*
* Reference:
* http://blog.csdn.net/linhuanmars/article/details/22839819
*
* Analysis from the above blog:
* 这道题是树的题目，属于最基本的树遍历的问题。问题要求就是判断两个树是不是一样，
* 基于先序，中序或者后序遍历都可以做完成，因为对遍历顺序没有要求。
*
* 这里我们主要考虑一下结束条件，如果两个结点都是null，也就是到头了，那么返回true。
* 如果其中一个是null，说明在一棵树上结点到头，另一棵树结点还没结束，即树不相同，
* 或者两个结点都非空，并且结点值不相同，返回false。
*
* 最后递归处理两个结点的左右子树，返回左右子树递归的与结果即可。
* 这里使用的是先序遍历，算法的复杂度跟遍历是一致的，如果使用递归，时间复杂度是O(n)，
* 空间复杂度是O(logn)
*
* Julia's comment:
* 1. online judge:
* 54 / 54 test cases passed.
Status: Accepted
Runtime: 176 ms
*/
public static bool IsSameTree(TreeNode p, TreeNode q)
{
if (p == null && q == null) // both nodes reaches the end
return true;
if (p == null || q == null) // only one of them reaches the end
return false;
if (p.val != q.val) // compare the value
return false;
return IsSameTree(p.left, q.left) && IsSameTree(p.right, q.right); // two subtrees comparison
}
/*
* Reference:
* http://www.cnblogs.com/lautsie/p/3247097.html
*
* julia's comment:
* 1. online judge:
* 54 / 54 test cases passed.
Status: Accepted
Runtime: 156 ms
*
*/
public static bool IsSameTree_B(TreeNode p, TreeNode q)
{
if (p == null && q == null) return true;
if (p != null && q != null)
{
if (p.val == q.val &&
IsSameTree_B(p.left, q.left) &&
IsSameTree_B(p.right, q.right))
return true;
}
return false;
}
/*
* Reference:
* http://www.cnblogs.com/lautsie/p/3247097.html
* comment from the above blog:
* 非递归版本，使用了Queue，有点类似BFS。顺便说下Java里面的Queue真难用，
* 连个empty()都没有，要用LinkedList（继承于Queue）
*
* http://blog.csdn.net/sunbaigui/article/details/8981275
* Analysis from the above blog:
*
* julia's comment:
*
* 1. Read through C# LinkedList methods
* 2. Try to find analog of Java LinkedList poll method in C#:
* First,
* RemoveFirst()
* 3. First time, use LinkedListNode class in C#
* LinkedList
* LinkedListNode
* understand why there are two classes, difference from Java - later.
* 4. Java LinkedList offer - add the tail of list vs C# LinkedList AddLast
* 5. online judge:
* 54 / 54 test cases passed.
Status: Accepted
Runtime: 168 ms
* 6. The tree traversal is kind of level traversal, BFS, two trees are compared
* to level by level;
* Julia argues that DFS search traversal does not work, counter example:
* two tree can be different, but in-order traversal can be the same.
* 2 3
* / \ / \
* 1 4 <- in order traversal both 1 2 3 4 - > 2 4
* / /
* 3 1
*/
public static bool isSameTree(TreeNode p, TreeNode q)
{
LinkedList<TreeNode> left = new LinkedList<TreeNode>();
LinkedList<TreeNode> right = new LinkedList<TreeNode>();
left.AddFirst(p); // java LinkedList offer vs C# LinkedList AddFirst ?
right.AddFirst(q);
while (left.Count > 0 && right.Count > 0)
{
LinkedListNode<TreeNode> l_f = left.First;
LinkedListNode<TreeNode> r_f = right.First;
left.RemoveFirst();
right.RemoveFirst();
TreeNode ln = l_f.Value;
TreeNode rn = r_f.Value;
if (ln == null && rn == null)
continue;
if (ln == null || rn == null)
return false;
if (ln.val != rn.val)
return false;
left.AddLast(ln.left);
left.AddLast(ln.right);
right.AddLast(rn.left);
right.AddLast(rn.right);
}
if (left.Count > 0 || right.Count > 0)
return false;
return true;
}