Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

 

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

 

Note:
Bonus points if you could solve it both recursively and iteratively.

 

判断一棵树是否是对称的。

 

/**
 * 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;

    return getResult(root.left,root.right);
        
    }

    public boolean getResult(TreeNode left,TreeNode right){
        if( left == null && right == null)
            return true;
        if( left == null || right == null)
            return false;
        if( left.val != right.val )
            return false;
        if( getResult(left.left,right.right) )
            return getResult(left.right,right.left);
        return false;

    }

}

 

 也可以使用队列来解决这个问题。