Leetcode 669-修剪二叉搜索树

题目描述

https://leetcode.cn/problems/trim-a-binary-search-tree/description/

解题思路

对于每个节点，如果其值小于区间最小值，不能直接将其完全删除，而应判断其右子树的值（该节点值小于区间，但右子树的值大于该节点，可能属于区间范围内）；同理，如果节点值大于区间范围，则需要进一步判断该节点的左子树是否属于区间范围。

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

Leetcode 108-将有序数组转换为二叉搜索树

题目描述

https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/

解题思路

/**
 * 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){
        //确定切分区间的开闭
        if (left>right) return nullptr;//左闭右闭，如果左大于右，则说明当前区间不合法，返回空节点
        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) {
        return traversal(nums,0, nums.size()-1);
    }
};

Leetcode 538-把二叉搜索树转换为累加树

题目描述

https://leetcode.cn/problems/convert-bst-to-greater-tree/description/

解题思路

根据二叉搜索树的性质可知，右侧节点大于根节点大于左侧节点，因此采用右中左的遍历顺序，通过 pre 和 cur 双指针对每个节点的值进行累加

/**
 * 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;
    TreeNode* convertBST(TreeNode* root) {
        if(root == nullptr) return nullptr;//终止条件
        root ->right = convertBST(root->right);
        root->val += pre;
        pre = root->val;
        root->left = convertBST(root->left);
        return root;
    }
};