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面试题 28. 对称的二叉树

题目描述

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/

解法

Python3

# 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 is_symmetric(left, right):
            if left is None and right is None:
                return True
            if left is None or right is None or left.val != right.val:
                return False
            return is_symmetric(left.left, right.right) and is_symmetric(
                left.right, right.left
            )

        if root is None:
            return True
        return is_symmetric(root.left, root.right)

Java

/**
 * 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) {
        if (root == null) return true;
        return isSymmetric(root.left, root.right);
    }

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

JavaScript

/**
 * 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) {
    function dfs(left, right) {
        if (!left && !right) return true;
        if (!left || !right || left.val != right.val) return false;
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }
    if (!root) return true;
    return dfs(root.left, root.right);
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSymmetric(root *TreeNode) bool {
    if root == nil {
        return true
    }
    return isSymme(root.Left, root.Right)
}

func isSymme(left *TreeNode, right *TreeNode) bool {
    if left == nil && right == nil {
        return true
    }
    if left == nil || right == nil || left.Val != right.Val {
        return false
    }
    return isSymme(left.Left, right.Right) && isSymme(left.Right, right.Left)
}

C++

/**
 * 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* left, TreeNode* right) {
        // 均为空，则直接返回 true。有且仅有一个不为空，则返回 false
        if (left == nullptr && right == nullptr) {
            return true;
        }
        if (left == nullptr || right == nullptr || left->val != right->val) {
            return false;
        }
        return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);
    }

    bool isSymmetric(TreeNode* root) {
        if (root == nullptr) {
            return true;
        }

        return isSymmetric(root->left, root->right);
    }
};

TypeScript

/**
 * 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 {
    if (root == null) {
        return true;
    }
    const dfs = (left: TreeNode | null, right: TreeNode | null) => {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null || left.val != right.val) {
            return false;
        }
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    };
    return dfs(root.left, root.right);
}

Rust

// 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(left: &Option<Rc<RefCell<TreeNode>>>, right: &Option<Rc<RefCell<TreeNode>>>) -> bool {
        if left.is_none() && right.is_none() {
            return true;
        }
        if left.is_none() || right.is_none() {
            return false;
        }
        let l = left.as_ref().unwrap().borrow();
        let r = right.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 {
        if root.is_none() {
            return true;
        }
        let node = root.as_ref().unwrap().borrow();
        Self::dfs(&node.left, &node.right)
    }
}

C#

/**
 * 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) {
        if (root == null) {
            return true;
        }
        return dfs(root.left, root.right);
    }

    public bool dfs(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null || left.val != right.val) {
            return false;
        }
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }
}

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