LeetCode 101 Symmetric Tree

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本文介绍了一种检查二叉树是否为中心对称的方法，包括递归和迭代两种实现方式，并解释了二叉树的序列化过程。

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

But the following is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

1.采用递归。判断左右子树是否互为镜面。

	public boolean isSymmetric(TreeNode root) {
		if (root == null) return true;
		return mirror(root.left, root.right);
	}

	public boolean mirror(TreeNode p, TreeNode q) {
		if (p == null && q == null) return true;
		if (p == null || q == null) return false;
		return p.val == q.val && mirror(p.left, q.right) && mirror(p.right, q.left);
	}

2.使用stack，stack好一点是，如果元素为null，也照样可以push、pop。

代码如下：

	public boolean isSymmetric2(TreeNode root) {
		Stack<TreeNode> nodeStack = new Stack<TreeNode>();
		if (root == null) return true;
		nodeStack.push(root.left);
		nodeStack.push(root.right);
		while (!nodeStack.isEmpty()) {
			TreeNode q = nodeStack.pop();
			TreeNode p = nodeStack.pop();
			if (p == null && q == null) continue;
			if (p == null || q == null) return false;
			if (p.val != q.val) return false;
			nodeStack.push(p.left);
			nodeStack.push(q.right);
			nodeStack.push(p.right);
			nodeStack.push(q.left);
		}
		return true;
	}