Convert Sorted Array to Binary Search Tree问题及解法
2017-03-20 16:13
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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;
}
};
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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