剑指 Offer 28. 对称的二叉树 双百解法，简单易懂二叉树镜像翻转对称

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
    	//判断根结点是否为空或递归调用先序遍历
        return root == null || preorder(root.left, root.right);
    }
    public boolean preorder(TreeNode root,TreeNode root2){
        if(root == null&&root2==null) return true;
        if(root==null||root2==null||root.val!=root2.val) return false;
		//注意这一块因为是镜像，所以root的左节点要和root2的右节点进行比较
		//root的右节点要和root2的左节点进行比较
        return preorder(root.left,root2.right)&&preorder(root.right,root2.left);
    }
}