数组和链表关于BST的转换

algorithm

数组转换成BST（二叉搜索树）

• leetcode题目地址108. Convert Sorted Array to Binary Search Tree

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

• 时间复杂度:$O(n)$，$T(n) = 2T(n/2) + c$
• 空间复杂度:$O(log(n))$,需要用到栈

链表转换为二叉搜索树

• leetcode地址,109. Convert Sorted List to Binary Search Tree

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * 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 {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        vector<ListNode*> vec;
        while(head) {
            vec.push_back(head);
            head = head->next;
        }
        int n = vec.size();
        return func(vec,0,n-1);
    }
    TreeNode* func(vector<ListNode*> &nums,int i ,int j) {
        if(i > j )
            return nullptr;
        if(i == j ) {
            return new TreeNode(nums[i]->val);
        }
        int mid = (i + j) / 2;
        TreeNode *root = new TreeNode(nums[mid]->val);
        root->left = func(nums,i,mid-1);
        root->right = func(nums,mid+1,j);
        return root;
    }
};

• 时间复杂度:$O(n)$，$T(n) = 2T(n/2) + c$
• 空间复杂度:$O(O(n))$,需要用到额外的数组