LeetCode 101 Symmetric Tree

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

For example, this binary tree is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following is not:

    1   / \  2   2   \   \   3    3

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:

   1  / \ 2   3    /   4    \     5

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;}