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;
}