| comments | difficulty | edit_url |
|---|---|---|
true |
简单 |
请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样，那么它是对称的。
例如，二叉树 [1,2,2,3,4,4,3] 是对称的。
1
/ \
2 2
/ \ / \
3 4 4 3
但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
1
/ \
2 2
\ \
3 3
示例 1：
输入：root = [1,2,2,3,4,4,3] 输出：true
示例 2：
输入：root = [1,2,2,null,3,null,3] 输出：false
限制：
0 <= 节点个数 <= 1000
注意：本题与主站 101 题相同：https://leetcode.cn/problems/symmetric-tree/
我们设计一个递归函数 dfs，它接收两个参数 a 和 b，分别代表两棵树的根节点。我们可以对 a 和 b 进行如下判断：
- 如果
a和b都为空，说明两棵树都遍历完了，返回true； - 如果
a和b中有且只有一个为空，说明两棵树的结构不同，返回false； - 如果
a和b的值不相等，说明两棵树的结构不同，返回false； - 如果
a和b的值相等，那么我们分别递归地判断a的左子树和b的右子树，以及a的右子树和b的左子树是否对称。
最后，我们返回 dfs(root, root)，即判断 root 的左子树和右子树是否对称。
时间复杂度
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
def dfs(a, b):
if a is None and b is None:
return True
if a is None or b is None or a.val != b.val:
return False
return dfs(a.left, b.right) and dfs(a.right, b.left)
return dfs(root, root)/**
* 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) {
return dfs(root, root);
}
private boolean dfs(TreeNode a, TreeNode b) {
if (a == null && b == null) {
return true;
}
if (a == null || b == null || a.val != b.val) {
return false;
}
return dfs(a.left, b.right) && dfs(a.right, b.left);
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
function<bool(TreeNode*, TreeNode*)> dfs = [&](TreeNode* a, TreeNode* b) -> bool {
if (!a && !b) {
return true;
}
if (!a || !b || a->val != b->val) {
return false;
}
return dfs(a->left, b->right) && dfs(a->right, b->left);
};
return dfs(root, root);
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isSymmetric(root *TreeNode) bool {
var dfs func(a, b *TreeNode) bool
dfs = func(a, b *TreeNode) bool {
if a == nil && b == nil {
return true
}
if a == nil || b == nil || a.Val != b.Val {
return false
}
return dfs(a.Left, b.Right) && dfs(a.Right, b.Left)
}
return dfs(root, root)
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function isSymmetric(root: TreeNode | null): boolean {
const dfs = (a: TreeNode | null, b: TreeNode | null): boolean => {
if (!a && !b) {
return true;
}
if (!a || !b || a.val != b.val) {
return false;
}
return dfs(a.left, b.right) && dfs(a.right, b.left);
};
return dfs(root, root);
}// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(a: &Option<Rc<RefCell<TreeNode>>>, b: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if a.is_none() && b.is_none() {
return true;
}
if a.is_none() || b.is_none() {
return false;
}
let l = a.as_ref().unwrap().borrow();
let r = b.as_ref().unwrap().borrow();
l.val == r.val && Self::dfs(&l.left, &r.right) && Self::dfs(&l.right, &r.left)
}
pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
Self::dfs(&root, &root)
}
}/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isSymmetric = function (root) {
const dfs = (a, b) => {
if (!a && !b) {
return true;
}
if (!a || !b || a.val != b.val) {
return false;
}
return dfs(a.left, b.right) && dfs(a.right, b.left);
};
return dfs(root, root);
};/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public bool IsSymmetric(TreeNode root) {
return dfs(root, root);
}
private bool dfs(TreeNode a, TreeNode b) {
if (a == null && b == null) {
return true;
}
if (a == null || b == null || a.val != b.val) {
return false;
}
return dfs(a.left, b.right) && dfs(a.right, b.left);
}
}/* public class TreeNode {
* var val: Int
* var left: TreeNode?
* var right: TreeNode?
* init(_ val: Int) {
* self.val = val
* self.left = nil
* self.right = nil
* }
* }
*/
class Solution {
func isSymmetric(_ root: TreeNode?) -> Bool {
return dfs(root, root)
}
private func dfs(_ a: TreeNode?, _ b: TreeNode?) -> Bool {
if a == nil && b == nil {
return true
}
if a == nil || b == nil || a!.val != b!.val {
return false
}
return dfs(a!.left, b!.right) && dfs(a!.right, b!.left)
}
}