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剑指offer-28

题目

结果

代码

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class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return isSymmetric(root.left, root.right);
}

private boolean isSymmetric(TreeNode left, TreeNode right) {
if (left == null || right == null) {
return left == right;
}
return left.val == right.val && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
}

复杂度

时间复杂度:O(n),n为节点个数

空间复杂度:O(递归层数)