Symmetric Tree leetcode java

题目：

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.

题解：

判断左右子树是否相等。

递归的解法：

 1  public boolean isSymmetricTree(TreeNode p,TreeNode q){
 2      if(p == null&&q == null)
 3         return true;
 4      if(p == null||q == null)
 5         return false;
 6      return (p.val == q.val) && isSymmetricTree(p.left, q.right) && isSymmetricTree(p.right, q.left);
 7 }
 8 
 9 public boolean isSymmetric(TreeNode root) {
10     if(root==null) 
11         return true;
12         
13     return isSymmetricTree(root.left,root.right);
14 }

非递归解法：

 1 public boolean isSymmetric(TreeNode root) {
 2     if(root == null)
 3         return true;
 4     if(root.left == null && root.right == null)
 5         return true;
 6     if(root.left == null || root.right == null)
 7         return false;
 8     LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();
 9     LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();
10     q1.add(root.left);
11     q2.add(root.right);
12     while(!q1.isEmpty() && !q2.isEmpty()){
13         TreeNode n1 = q1.poll();
14         TreeNode n2 = q2.poll();
15         
16         if(n1.val != n2.val)
17             return false;
18         if((n1.left == null && n2.right != null) || (n1.left != null && n2.right == null))
19             return false;
20         if((n1.right == null && n2.left != null) || (n1.right != null && n2.left == null))
21             return false;
22         
23         if(n1.left != null && n2.right != null){
24             q1.add(n1.left);
25             q2.add(n2.right);
26         }
27         
28         if(n1.right != null && n2.left != null){
29             q1.add(n1.right);
30             q2.add(n2.left);
31         }            
32     }
33     return true;
34 }

 Reference：

 http://blog.csdn.net/linhuanmars/article/details/23072829