面试必刷TOP101：31、对称的二叉树

/*思路：首先根节点以及其左右子树，左子树的左子树和右子树的右子树相同
* 左子树的右子树和右子树的左子树相同即可，采用递归
* 非递归也可，采用栈或队列存取各级子树根节点
*/
public class Solution {
	boolean isSymmetrical(TreeNode pRoot)
	{
		if(pRoot == null){
			return true;
		}
		return comRoot(pRoot.left, pRoot.ri服务器托管网ght);
	}
	private boolean comRoot(TreeNode left, TreeNode right) {
		// TODO Auto-generated method stub
		if(left == null) return right==null;
		if(right == null) return false;
		if(left.val != right.val) return false;
		return comRoot(left.right, right.left) && comR服务器托管网oot(left.left, right.right);
	}
}