101. 对称二叉树 

题目描述

给你一个二叉树的根节点 root ，检查它是否轴对称。

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

提示：

树中节点数目在范围 [1, 1000] 内
-100 <= Node.val <= 100

进阶：你可以运用递归和迭代两种方法解决这个问题吗？

解法

方法一：递归

我们设计一个函数 $\textit{dfs}(\textit{root1}, \textit{root2})$，用于判断两个二叉树是否对称。答案即为 $\textit{dfs}(\textit{root.left}, \textit{root.right})$。

函数 $\textit{dfs}(\textit{root1}, \textit{root2})$ 的逻辑如下：

如果 $\textit{root1}$ 和 $\textit{root2}$ 都为空，则两个二叉树对称，返回 true；
如果 $\textit{root1}$ 和 $\textit{root2}$ 中只有一个为空，或者 $\textit{root1.val} \neq \textit{root2.val}$
否则，判断 $\textit{root1}$ 的左子树和 $\textit{root2}$ 的右子树是否对称，以及 $\textit{root1}$ 的右子树和 $\textit{root2}$ 的左子树是否对称，这里使用了递归。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        def dfs(root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
            if root1 == root2:
                return True
            if root1 is None or root2 is None or root1.val != root2.val:
                return False
            return dfs(root1.left, root2.right) and dfs(root1.right, root2.left)

        return dfs(root.left, root.right)

Java

/**
 * 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.left, root.right);
    }

    private boolean dfs(TreeNode root1, TreeNode root2) {
        if (root1 == root2) {
            return true;
        }
        if (root1 == null || root2 == null || root1.val != root2.val) {
            return false;
        }
        return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        auto dfs = [&](this auto&& dfs, TreeNode* root1, TreeNode* root2) -> bool {
            if (root1 == root2) {
                return true;
            }
            if (!root1 || !root2 || root1->val != root2->val) {
                return false;
            }
            return dfs(root1->left, root2->right) && dfs(root1->right, root2->left);
        };
        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 {
	var dfs func(root1, root2 *TreeNode) bool
	dfs = func(root1, root2 *TreeNode) bool {
		if root1 == root2 {
			return true
		}
		if root1 == nil || root2 == nil || root1.Val != root2.Val {
			return false
		}
		return dfs(root1.Left, root2.Right) && dfs(root1.Right, root2.Left)
	}
	return dfs(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 {
    const dfs = (root1: TreeNode | null, root2: TreeNode | null): boolean => {
        if (root1 === root2) {
            return true;
        }
        if (!root1 || !root2 || root1.val !== root2.val) {
            return false;
        }
        return dfs(root1.left, root2.right) && dfs(root1.right, root2.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::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        fn dfs(root1: Option<Rc<RefCell<TreeNode>>>, root2: Option<Rc<RefCell<TreeNode>>>) -> bool {
            match (root1, root2) {
                (Some(node1), Some(node2)) => {
                    let node1 = node1.borrow();
                    let node2 = node2.borrow();
                    node1.val == node2.val
                        && dfs(node1.left.clone(), node2.right.clone())
                        && dfs(node1.right.clone(), node2.left.clone())
                }
                (None, None) => true,
                _ => false,
            }
        }

        match root {
            Some(root) => dfs(root.borrow().left.clone(), root.borrow().right.clone()),
            None => true,
        }
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function (root) {
    const dfs = (root1, root2) => {
        if (root1 === root2) {
            return true;
        }
        if (!root1 || !root2 || root1.val !== root2.val) {
            return false;
        }
        return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
    };
    return dfs(root.left, root.right);
};