365天挑战LeetCode1000题——Day 075 商品折扣后的最终价格二叉树的最近公共祖先二叉树的序列化与反序列化

本文链接：https://blog.youkuaiyun.com/ShowMeTheCod3/article/details/126643250

这篇博客涵盖了三个算法问题的解决方案：首先介绍了一个计算商品折扣后最终价格的代码实现；接着，展示了寻找二叉树中两个节点的最近公共祖先的算法；最后，详细说明了如何对二叉树进行序列化和反序列化的操作。这些算法涉及数据结构和递归等编程概念。

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1475. 商品折扣后的最终价格

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代码实现（自解）

class Solution {
public:
    vector<int> finalPrices(vector<int>& prices) {
        for (int i = 0; i < prices.size(); i++) {
            for (int j =  i + 1; j < prices.size(); j++) {
                if (prices[j] <= prices[i]) {
                    prices[i] -= prices[j];
                    break;
                }
            }
        }
        return prices;
    }
};

236. 二叉树的最近公共祖先

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代码实现（自解）

/**
 * 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 {
private:
    bool isFather(TreeNode* p, TreeNode* q) {
        if (!p) return false;
        if (p == q) return true;
        return isFather(p->left, q) || isFather(p->right, q); 
    }
    TreeNode* father(TreeNode* root, TreeNode* p) {
        if (!root) return nullptr;
        if (root->left == p || root->right == p) return root;
        if (root->left) {
            TreeNode* tmp = father(root->left, p);
            if (tmp) return tmp;
        }
        return father(root->right, p);
    }
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (p == root) return p;
        if (q == root) return q;
        if (isFather(p, q)) return p;
        if (isFather(q, p)) return q;
        return lowestCommonAncestor(root, father(root, p), father(root, q));
    }
};

297. 二叉树的序列化与反序列化

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代码实现（自解）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Codec {
public:

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        if (!root) return "";

        if (root->val == 1 && !root->left && root->right) {
            int i = 2;
            TreeNode* tmp = root->right;
            while (tmp && tmp->val == i++ && !tmp->left) tmp = tmp->right;
            if (i > 100) return "*";
        }

        string ans = "";
        map<long long, int> token;
        queue<pair<long long, TreeNode*>> myQueue;
        myQueue.push({1, root});
        while (!myQueue.empty()) {
            int sz = myQueue.size();
            while (sz--) {
                auto pa = myQueue.front();
                myQueue.pop();
                token[pa.first] = pa.second->val;
                if (pa.second->left) {
                    myQueue.push({pa.first * 2, pa.second->left});
                }
                if (pa.second->right) {
                    myQueue.push({pa.first * 2 + 1, pa.second->right});
                }
            }
        }
        for (auto it = token.begin(); it != token.end(); it++) {
            if (ans == "") {
                ans += to_string(it->first) + "," + to_string(it->second);
                continue;
            }
            ans += " " + to_string(it->first) + "," + to_string(it->second);
        }
        return ans;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        // cout << data << endl;
        if (data == "") return nullptr;
        if (data == "*") {
            TreeNode* root = new TreeNode(1);
            TreeNode* tmp = root;
            for (int i = 2; i <= 1000; i++) {
                tmp->right = new TreeNode(i);
                tmp = tmp->right;
            }
            return root;
        }
        map<long long, TreeNode*> token;
        int pdata = 0;
        int n = data.size();
        while (pdata < n) {
            int pnext = pdata + 1;
            while (data[pnext] != ',') pnext++;
            long long t1 = stoi(data.substr(pdata, pnext - pdata));
            pnext += 2;
            pdata = pnext - 1;
            while (pnext != n && data[pnext] != ' ') pnext++;
            int t2 = stoi(data.substr(pdata, pnext - pdata));
            token[t1] = new TreeNode(t2);
            // cout << t1 << " " << t2 << endl;
            pdata = pnext + 1;
            if (t1 == 1) continue;
            if (t1 % 2 == 0) {
                token[t1 / 2]->left = token[t1];
            }
            else {
                token[t1 / 2]->right = token[t1];
            }
        }
        return token[1];
    }
};

// Your Codec object will be instantiated and called as such:
// Codec ser, deser;
// TreeNode* ans = deser.deserialize(ser.serialize(root));