LeetCode:Convert Sorted Array to Binary Search Tree,Convert Sorted List to Binary Search Tree
LeetCode:Convert Sorted Array to Binary Search Tree
Given an array where elements are sorted in ascending order, convert it to a height balanced BST
分析：找到数组的中间数据作为根节点，小于中间数据的数组来构造作为左子树，大于中间数据的数组来构造右子树，递归解法如下
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode *sortedArrayToBST(vector<int> &num) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 int len = num.size(); 16 if(len == 0)return NULL; 17 return sortedArrayToBSTRecur(num, 0, len-1); 18 } 19 TreeNode *sortedArrayToBSTRecur(vector<int> &num, int istart, int iend) 20 { 21 if(istart > iend)return NULL; 22 int middle = (istart+iend)/2; 23 TreeNode *res = new TreeNode(num[middle]); 24 res->left = sortedArrayToBSTRecur(num, istart, middle-1); 25 res->right = sortedArrayToBSTRecur(num, middle+1, iend); 26 return res; 27 } 28 };
LeetCode: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
分析：和上一题同理，只不过要使用快慢指针来找到链表的中间节点 本文地址
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 /** 10 * Definition for binary tree 11 * struct TreeNode { 12 * int val; 13 * TreeNode *left; 14 * TreeNode *right; 15 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 16 * }; 17 */ 18 class Solution { 19 public: 20 TreeNode *sortedListToBST(ListNode *head) { 21 // IMPORTANT: Please reset any member data you declared, as 22 // the same Solution instance will be reused for each test case. 23 if(head == NULL)return NULL; 24 ListNode *fast = head, *slow = head, *preSlow = NULL; 25 while(fast->next && fast->next->next) 26 { 27 fast = fast->next->next; 28 preSlow = slow; 29 slow = slow->next; 30 } 31 TreeNode *res = new TreeNode(slow->val); 32 fast = slow->next; 33 delete slow; 34 if(preSlow != NULL) 35 { 36 preSlow->next = NULL; 37 res->left = sortedListToBST(head); 38 } 39 res->right = sortedListToBST(fast); 40 return res; 41 } 42 };
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