二叉树

226.翻转二叉树

Tag: 二叉树，广度优先搜索，深度优先搜索

题目

https://leetcode.cn/problems/invert-binary-tree/

Solution1

遍历二叉树，递归翻转

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/**
 * 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:
    TreeNode* invertTree(TreeNode* root) {
        if (root == nullptr) {
            return nullptr;
        }

        TreeNode *tmp = root->left;
        root->left = invertTree(root->right);
        root->right = invertTree(tmp);

        return root;
    }
};

101.对称二叉树

Tags: 二叉树，广度优先搜索，深度优先搜索

题目

https://leetcode.cn/problems/symmetric-tree/

Solution1

深度优先搜索，递归判断

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/**
 * 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) {}
 * };
 */
bool dfs(TreeNode *left, TreeNode *right) {
    // 判断非空
    if (left == nullptr || right == nullptr) {
        return ((left == nullptr) && (right == nullptr));
    }

    if (left->val != right->val) {
        return false;
    }

    // 递归
    return dfs(left->left, right->right) && dfs(left->right, right->left);
}

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (root == nullptr)
            return true;

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

  • 优点：简洁易懂

  • 缺点：没啥缺点，但是递归。可以优化为尾递归

  • 结论：好办法

链表

92.反转链表 II

Tags: 链表

题目

https://leetcode.cn/problems/reverse-linked-list-ii/

Solution1