Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. 思路:题目看上去好像很难,但实际上很简单,递归做就行,每次找到左右子树对应的子链表就行.一次AC. class Solution { public: TreeNode *sortedListToBST(ListNode *head) { if(head == NULL) retu…

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.     可以采用类似于Covert Sorted Array to Binary Search Tree的方法,但是寻找中点对于链表来说效率较低 可以采用更高效的递归方式,无需寻找中点 注意引用传…

Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in ascending order, convert it to a height balanced BST.   每次把中间元素当成根节点,递归即可   /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * Tre…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 有序数组变二叉平衡搜索树,不难,递归就行.每次先序建立根节点(取最中间的数),然后用子区间划分左右子树. 一次就AC了 注意:new 结构体的时候对于 struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x)…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.思路:对于树来说,自顶向下的递归是很好控制的,自底向上的递归就很容易让脑神经打结了.本来想仿照排序二叉树的中序遍历,逆向地由数组构造树,后来发现这样做需要先确定树的结构,不然会一直递归下去,不知道左树何时停止.代码: TreeNode *addNode(vector<int> &num, int…

和上题思路基本一致,不同的地方在于,链表不能随机访问中间元素. int listLength(ListNode* node) { ; while (node) { n++; node = node->next; } return n; } ListNode* nth_node(ListNode* node, int n) { while (--n)node = node->next; return node; } TreeNode* sortedListToBST(ListNode* head…

思路很简单,用二分法,每次选中间的点作为根结点,用左.右结点递归. TreeNode* sortedArrayToBST(vector<int> &num) { return sortedArrayToBST(num.begin(), num.end()); } template<typename RandomAccessIterator> TreeNode* sortedArrayToBST(RandomAccessIterator first, RandomAccess…

108. Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the tw…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 题目标签:Tree 这道题目给了我们一个有序数组,从小到大.让我们把这个数组转化为height balanced BST. 首先来看一下什么是binary search tree: 每一个点的left < 节点 < right, 换一句话说,每一个点的值要大于左边的,小于右边的. 那么什么是heigh…

Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 解题思路: 用一个有序数组,创建一个平衡二叉查找树. 为确保平衡,需要满足两子树的高度差不大于1,可以通过设置左子树结点数等于或者比右子树结点数多1,来实现. 那么每次取数组的中间位置后一个值,作为根结点,数组左边元素的插入左子树,数组右边元素插入右子树,依次类推. 代码: /** * Definiti…