subject ：

Please implement a function , Used to judge whether a binary tree is symmetrical . Be careful , If a binary tree has the same image as this binary tree , Define it as symmetric .

for example , Binary tree  [1,2,2,3,4,4,3] It's symmetrical .

    1
   / \
  2   2
 / \ / \
3  4 4  3
But the next one  [1,2,2,null,3,null,3] It's not mirror symmetric :

    1
   / \
  2   2
   \   \
   3    3

answer ：

Solution 1 ：

recursive , The code is as follows ：

/**
 * 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) {
        if(root == null){
            return true;
        }
        return compare(root.left, root.right);
    }
    public boolean compare(TreeNode A, TreeNode B){
        if(A == null && B == null){
            return true;
        }
        if(A == null || B == null){
            return false;
        }
        return A.val == B.val && compare(A.left, B.right) && compare(A.right, B.left);
    }
}

Solution 2 ：

iteration , The code is as follows ：

public class Solution {
    boolean isSymmetrical(TreeNode pRoot)
    {
        if(pRoot==null){
            return true;
        }
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(pRoot.left);
        queue.add(pRoot.right);
        while(!queue.isEmpty()){
            TreeNode node1 = queue.poll();
            TreeNode node2 = queue.poll();
            if(node1==null&&node2==null){
                continue;
            }
            if(node1==null||node2==null){
                return false;
            }
            if(node1.val==node2.val){
                queue.add(node1.left);
                queue.add(node2.right);
                queue.add(node1.right);
                queue.add(node2.left);
            }else{
                return false;
            }
        }
       return true;
    }
}

Reference link ：https://leetcode-cn.com/problems/symmetric-tree/solution/dui-cheng-er-cha-shu-by-leetcode/