image.png

解法

递归方法如下：

/**
 * 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 {
    public boolean isSymmetric(TreeNode root) {
        return dfs(root, root);
    }

    public boolean dfs(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null) {
            return false;
        }
        if (left.val != right.val) {
            return false;
        }
        // 左节点的左子树和右节点的右子树比较，递归一直拿到两者对应的节点
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }
}

迭代方法

/**
 * 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) {
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        q.offer(root);
        while (!q.isEmpty()) {
            TreeNode left = q.poll();
            TreeNode right = q.poll();
            if (left == null && right == null) {
                continue;
            }
            if (left == null || right == null) {
                return false;
            }
            if (left.val != right.val) {
                return false;
            }
            // 第一对节点入队列
            q.offer(left.left);
            q.offer(right.right);
            // 第二对节点入队列
            q.offer(left.right);
            q.offer(right.left);
        }
        return true;
    }
}