Convert Sorted List to Binary Search Tree——将链表转换为平衡二叉搜索树 &&convert-sorted-array-to-binary-search-tree——将数列转换为bst

Convert Sorted List to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

为了满足平衡要求，容易想到提出中间节点作为树根，因为已排序，所以左右两侧天然满足BST的要求。

左右子串分别递归下去，上层根节点连接下层根节点即可完成。

递归找中点，然后断开前后两段链表，并继续找中点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if(head==NULL) return NULL;
        if(head->next==NULL) return new TreeNode(head->val);
        ListNode *mid=findMid(head);
        TreeNode *root=new TreeNode(mid->val);
        root->left=sortedListToBST(head);
        root->right=sortedListToBST(mid->next);
        return root;
    }
    ListNode *findMid(ListNode *root){
        if(root==NULL) return NULL;
        if(root->next==NULL) return root;
        ListNode *fast,*slow,*pre;
        fast=root;
        slow=root;
        while(fast!=NULL){
            fast=fast->next;
            if(fast!=NULL){
                fast=fast->next;
                pre=slow;
                slow=slow->next;
            }
        }
        pre->next=NULL;
        return slow;
    }
};

convert-sorted-array-to-binary-search-tree

 Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedArrayToBST(vector<int> &num) {
        int n=num.size();
        if(n<1) return NULL;
        
        return findMid(num,0,n-1);
    }
    TreeNode *findMid(vector<int> &num, int l,int r){
        if(l>r) return NULL;
        int mid=(l+r+1)/2;
        TreeNode *root=new TreeNode(num[mid]);
        root->left=findMid(num,l,mid-1);
        root->right=findMid(num,mid+1,r);
        return root;
    }
};