/*
* Reference:
* http://www.cnblogs.com/TenosDoIt/p/3440729.html
*
* Analysis from the above blog:
* 算法1：递归解法，判断左右两颗子树是否对称，只要两颗子树的根节点值相同，
* 并且左边子树的左子树和右边子树的右子树对称 且左边子树的右子树和右边子
* 树的左子树对称
*
* julia's comment:
* 1. online judge:
* 192 / 192 test cases passed.
Status: Accepted
Runtime: 176 ms
*/
public static bool isSymmetric(TreeNode root)
{
if (root == null)
return true;
return isSymmetricRecur(root.left, root.right);
}
/**
*
*/
public static bool isSymmetricRecur(TreeNode r1, TreeNode r2)
{
if (r1 != null && r2 != null) // neither of two nodes is null
{
if (r1.val == r2.val &&
isSymmetricRecur(r1.left, r2.right) &&
isSymmetricRecur(r1.right, r2.left))
return true;
else
return false;
}
else if (r1 != null || r2 != null) // one of them is not null, but at least one is null
return false;
else // two of them are null
return true;
}
/*
* code reference:
* http://www.cnblogs.com/TenosDoIt/p/3440729.html
* Iterative solution:
算法2：非递归解法，用两个队列分别保存左子树节点和右子树节点，每次从两个队列中
* 分别取出元素，如果两个元素的值相等，则继续把他们的左右节点加入左右队列。要注意
* 每次取出的两个元素，左队列元素的左孩子要和右队列元素的右孩子要同时不为空或者
* 同时为空，否则不可能对称，同理左队列元素的右孩子要和右队列元素的左孩子也一样。
*
* julia's comment:
* 1. online judge:
192 / 192 test cases passed.
Status: Accepted
Runtime: 156 ms
* 2. The code can be improved by removing duplicated checking
* 1st node'left child vs 2nd node's right child
* 1st node'right child vs 2nd node's right child
*
* The rewrite version is called isSymmetric_Iterative_ShortVersion
*/
public static bool isSymmetric_Iterative(TreeNode root)
{
if (root == null)
return true;
Queue lQ = new Queue(),
rQ = new Queue();
if (root.left != null)
lQ.Enqueue(root.left);
if (root.right != null)
rQ.Enqueue(root.right);
while (lQ.Count > 0 && rQ.Count > 0)
{
TreeNode l_nd = (TreeNode)lQ.Peek(); // l_nd: left queue node
TreeNode r_nd = (TreeNode)rQ.Peek(); // r_nd: right queue node
lQ.Dequeue();
rQ.Dequeue();
if (l_nd.val == r_nd.val)
{
// l_nd (node from 1st queue) left child vs right queue (node from 2nd queue) right child
if (l_nd.left != null && r_nd.right != null) // neither is null
{
lQ.Enqueue(l_nd.left);
rQ.Enqueue(r_nd.right);
}
else if (l_nd.left != null || r_nd.right != null) // one of them is null, another is not null
return false;
// 1st queue's node's right child vs 2nd queue's node's left child
if (l_nd.right != null && r_nd.left != null)
{
lQ.Enqueue(r_nd.left);
rQ.Enqueue(l_nd.right);
}
else if (r_nd.left != null || l_nd.right != null)
return false;
}
else return false;
}
if (lQ.Count == 0 && rQ.Count == 0)
return true;
else
return false;
}
/*
* julia's comment:
* make iterative solution short
* 1. abstract code using two nodes: n1, n2
* 2. make a loop iteration two times.
* 3. think about how to make code more simple, readable, maintainable.
*
*/
public static bool isSymmetric_Iterative_ShortVersion(TreeNode root)
{
if (root == null)
return true;
Queue lQ = new Queue(),
rQ = new Queue();
if (root.left != null)
lQ.Enqueue(root.left);
if (root.right != null)
rQ.Enqueue(root.right);
while (lQ.Count > 0 && rQ.Count > 0)
{
TreeNode l_nd = (TreeNode)lQ.Peek(); // l_nd: left queue node
TreeNode r_nd = (TreeNode)rQ.Peek(); // r_nd: right queue node
lQ.Dequeue();
rQ.Dequeue();
if (l_nd.val == r_nd.val)
{
for (int i = 0; i < 2; i++)
{
TreeNode n1 = l_nd.left;
TreeNode n2 = r_nd.right;
if (i == 1)
{
n1 = l_nd.right;
n2 = r_nd.left;
}
if (n1 != null && n2 != null) // neither is null
{
lQ.Enqueue(n1);
rQ.Enqueue(n2);
}
else if (n1 != null || n2 != null) // one of them is null, another is not null
return false;
}
}
else
return false;
}
if (lQ.Count == 0 && rQ.Count == 0) // both two queues are empty
return true;
else
return false;
}