最开始肯定是比较简单的想法，就是遍历左右节点呀，不相等我就直接返回false。

但是这样错了，我们要的是以根节点为轴，而不是以各个子节点。

反例：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        
        isSymmetric(root.left);
        isSymmetric(root.right);
        if(root.left == root.right) return true;
        else return false;
    }
}

正确做法

• 先排除不能遍历子节点的情况：

 // 首先排除空节点的情况
        if (left == null && right != null) return false;
        else if (left != null && right == null) return false;
        else if (left == null && right == null) return true;
        // 排除了空节点，再排除数值不相同的情况
        else if (left.val != right.val) return false;

• 再去递归子节点

完整代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    boolean compare(TreeNode left, TreeNode right){
        // 首先排除空节点的情况
        if (left == null && right != null) return false;
        else if (left != null && right == null) return false;
        else if (left == null && right == null) return true;
        // 排除了空节点，再排除数值不相同的情况
        else if (left.val != right.val) return false;

        // 如果相同，继续递归逻辑
        compare(left.left, right.right);
        compare(left.right, right.left);
        return compare(left.left, right.right) && compare(left.right, right.left);
    }
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return compare(root.left, root.right);

    }
}