Leetcode: 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.

难度：82. 这道题是树的题目，本质上还是树的遍历。这里无所谓哪种遍历方式，只需要对相应结点进行比较即可。一颗树对称其实就是看左右子树是否对称，一句话就是左同右，右同左，结点是对称的相等。不对称的条件有以下三个：（1）左边为空而右边不为空；（2）左边不为空而右边为空；（3）左边值不等于右边值。根据这几个条件在遍历时进行判断即可。

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
return check(root.left, root.right);
}

public boolean check(TreeNode node1, TreeNode node2) {
if (node1 == null && node2 == null) return true;
else if (node1 == null && node2 != null) return false;
else if (node1 != null && node2 == null) return false;
else if (node1.val != node2.val) return false;
return check(node1.left, node2.right) && check(node1.right, node2.left);
}
}

Iterative的方式参考了网上的解法：

public boolean isSymmetric(TreeNode root) {
if(root == null)
return true;
if(root.left == null && root.right == null)
return true;
if(root.left == null || root.right == null)
return false;
LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();
LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();
q1.add(root.left);
q2.add(root.right);
while(!q1.isEmpty() && !q2.isEmpty())
{
TreeNode n1 = q1.poll();
TreeNode n2 = q2.poll();

if(n1.val != n2.val)
return false;
if(n1.left == null && n2.right != null || n1.left != null && n2.right == null)
return false;
if(n1.right == null && n2.left != null || n1.right != null && n2.left == null)
return false;
if(n1.left != null && n2.right != null)
{
q1.add(n1.left);
q2.add(n2.right);
}
if(n1.right != null && n2.left != null)
{
q1.add(n1.right);
q2.add(n2.left);
}
}
return true;
}