「程式碼隨想錄演算法訓練營」第十八天 | 二叉樹 part8

669. 修剪二叉搜尋樹

題目連結：https://leetcode.cn/problems/trim-a-binary-search-tree/
題目難度：中等
文章講解：https://programmercarl.com/0669.修剪二叉搜尋樹.html
影片講解：https://www.bilibili.com/video/BV17P41177ud?share_source=copy_web
題目狀態：沒有思路，看題解過

思路：

使用遞迴。

1. 引數和返回值：引數是節點，和左右邊界，返回值是節點。
2. 終止條件：當節點為nullptr時，終止。
3. 單層迴圈：噹噹前節點的值小於最小邊界時，將其右孩子加入二叉樹中進行遍歷；噹噹前節點的值大於最大邊界時，將其左孩子加入二叉樹中進行遍歷。

程式碼實現：

/**
 * 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* trimBST(TreeNode* root, int low, int high) {
        if(root == nullptr) return nullptr;
        if(root->val < low) {
            TreeNode *right = trimBST(root->right, low, high);
            return right;
        }
        if(root->val > high) {
            TreeNode *left = trimBST(root->left, low, high);
            return left;
        }
        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);
        return root;
    }
};

迭代法程式碼：

/**
 * 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* trimBST(TreeNode* root, int low, int high) {
        if(root == nullptr) return nullptr;
        while(root != nullptr && (root->val < low || root->val > high)) {
            if(root->val < low) root = root->right;
            else root = root->left;
        }
        TreeNode *cur = root;
        while(cur != nullptr) {
            while(cur->left && cur->left->val < low) cur->left = cur->left->right;
            cur = cur->left;
        }
        cur = root;
        while(cur != nullptr) {
            while(cur->right && cur->right->val > high) cur->right = cur->right->left;
            cur = cur->right;
        }
        return root;
    }
};

108. 將有序陣列轉換為二叉搜尋樹

題目連結：https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/
題目難度：簡單
文章講解：https://programmercarl.com/0108.將有序陣列轉換為二叉搜尋樹.html
影片講解：https://www.bilibili.com/video/BV1uR4y1X7qL?share_source=copy_web
題目狀態：依舊是看題解

思路：

以陣列的中間為根結點，分別擷取陣列的左半部分作為根結點的左孩子，擷取陣列的右半部分作為根結點的右孩子。

程式碼：

/**
 * 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* traversal(vector<int>& nums, int left, int right) {
        if(left > right) return nullptr;
        int mid = left + (right - left) / 2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = traversal(nums, left, mid - 1);
        root->right = traversal(nums, mid + 1, right);
        return root;
    }
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode *root = traversal(nums, 0, nums.size() - 1);
        return root;
    }
};

538. 把二叉搜尋樹轉換為累加樹

題目連結：https://leetcode.cn/problems/convert-bst-to-greater-tree/
題目難度：中等
文章講解：https://programmercarl.com/0538.把二叉搜尋樹轉換為累加樹.html
影片講解：https://www.bilibili.com/video/BV1d44y1f7wP?share_source=copy_web
題目狀態：依舊是看題解

思路：

透過一個int型別的全域性變數來儲存比當前節點大的節點的值，並使用右中左的順序來遍歷二叉樹。

程式碼：

/**
 * 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:
    int pre = 0;
    void traversal(TreeNode* cur) {
        if(cur == nullptr) return;
        traversal(cur->right);
        cur->val += pre;
        pre = cur->val;
        traversal(cur->left);
    }
    TreeNode* convertBST(TreeNode* root) {
        pre = 0;
        traversal(root);
        return root;
    }
};

迭代法：

/**
 * 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:
    int pre;
    void traversal(TreeNode* root) {
        stack<TreeNode*> st;
        TreeNode* cur = root;
        while(cur != nullptr || !st.empty()) {
            if(cur != nullptr) {
                st.push(cur);
                cur = cur->right;
            } else {
                cur = st.top();
                st.pop();
                cur->val += pre;
                pre = cur->val;
                cur = cur->left;
            }
        }
    }
    TreeNode* convertBST(TreeNode* root) {
        pre = 0;
        traversal(root);
        return root;
    }
};