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

/**
 * Definition for a binary tree node.
 * 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;
        }
        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;
    }
}

?

Symmetric Tree

原文：http://hcx2013.iteye.com/blog/2237449