Convert Sorted Array to Binary Search Tree问题及解法
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问题描述:
Given an array where elements are sorted in ascending order, convert it to a height balanced BST。
问题分析:
将一个有序数组转化为平衡二叉搜索树。平衡二叉树有个特点:它的左子树和右子树均为平衡二叉树。因此可以考虑利用递归的方式求解。
先求根节点,再求左右子树。
过程详见代码:
/** * 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) { if(nums.size() == 0) return NULL; if(nums.size() == 1) return new TreeNode(nums[0]); int mid = nums.size() / 2; TreeNode* root = new TreeNode(nums[mid]); vector<int> leftInts(nums.begin(), nums.begin()+mid); vector<int> rightInts(nums.begin()+mid+1, nums.end()); root->left = sortedArrayToBST(leftInts); root->right = sortedArrayToBST(rightInts); return root; }};
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