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);}
}