#101 Symmetric Tree

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.

递归方法的代码如下：

/**
* 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;
return isSymmetricCompare(root.left, root.right);
}
public boolean isSymmetricCompare(TreeNode l, TreeNode r){
if(l == null && r ==null)
return true;
if(l == null || r ==null)//这句很漂亮,涵盖了几种情况！
return false;
return (l.val == r.val) && isSymmetricTree(l.left, ri.right) && isSymmetricTree(l.right, r.left);
}
}

非递归解法(用的是层序遍历)：

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);//java.util.LinkedList.add()将指定元素e追加到此列表的<u><strong>末尾</strong></u>
        q2.add(root.right);
        while(!q1.isEmpty() && !q2.isEmpty())
        {
            TreeNode n1 = q1.poll();//java.util.LinkedList.poll()返回列表的头元素，或null--如果此列表为空.调用以后会删除该元素
            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;
    }
}

写在最后：开始对非递归解法的思考在使用栈来解题。懒人思维没往下多想。有机会补上吧。

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