Convert Sorted Array to Binary Search Tree - LeetCode 108
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题目描述:
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
Hide Tags Tree Depth-first Search
首先,整颗数的节点的下标范围是[left,right],初始化分别为0,n-1,其中为n数组大小,并且根节点必然是数组中间节点v[mid]。
左子树的节点下标范围是[left,mid-1],右子树的节点下标范围是[mid+1,right]。
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
Hide Tags Tree Depth-first Search
分析:
要构造平衡的二叉搜索树(AVL),必须满足平衡条件:它是一棵空树或它的左右两个子树的高度差的绝对值不超过1,并且左右两个子树都是一棵平衡二叉树。
给定一个有序序列,可以按照以下方式递归构造AVL数,每次先构造出根节点,然后利用数组的连续序列递归构造出左子树和右子树:首先,整颗数的节点的下标范围是[left,right],初始化分别为0,n-1,其中为n数组大小,并且根节点必然是数组中间节点v[mid]。
左子树的节点下标范围是[left,mid-1],右子树的节点下标范围是[mid+1,right]。
以下是C++实现代码:
/**////////20ms//*/ * * 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* sortedArrayToBSTHelp(vector<int>& nums,int left,int right) { if(left <= right) { int mid = (left + right)/2; TreeNode* root = new TreeNode(nums[mid]); //以中间节点为根 root->left = sortedArrayToBSTHelp(nums,left,mid-1); //左边节点构造左子树 root->right = sortedArrayToBSTHelp(nums,mid+1,right); //右边节点构造右子树 return root; } else return NULL; } TreeNode* sortedArrayToBST(vector<int>& nums) { int size = nums.size(); if(size == 0) return NULL; return sortedArrayToBSTHelp(nums,0,size-1); }};
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