2023.9.8
本题分为两步，先用中序遍历将二叉搜索树转化为排序数组，再通过排序数组构建一个平衡二叉树。 代码如下：文章来源：https://www.toymoban.com/news/detail-702946.html
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
vector<int> v;
//将排序树转化为排序数组
void trans(TreeNode* cur)
{
if(cur == nullptr) return;
trans(cur->left);
v.push_back(cur->val);
trans(cur->right);
}
//将有序数组转化为平衡二叉树
TreeNode* getTree(vector<int> &nums , int left , int right)
{
if(left > right) return nullptr;
int seg = (left + right) / 2;
TreeNode* root = new TreeNode(nums[seg]);
root->left = getTree(nums , left , seg-1);
root->right = getTree(nums , seg+1 , right);
return root;
}
public:
TreeNode* balanceBST(TreeNode* root) {
trans(root);
TreeNode* node = getTree(v,0,v.size()-1);
return node;
}
};本题的这两个步骤可以当作模板背下来。文章来源地址https://www.toymoban.com/news/detail-702946.html
到了这里，关于leetcode 1382. 将二叉搜索树变平衡的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！
