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1382.balanceBST.cpp
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executable file
·59 lines (54 loc) · 1.5 KB
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/*
@filename 1382.balanceBST.cpp
@author caonan
@date 2022-04-23 13:29:44
@reference leetcode
@url https://leetcode-cn.com/problems/balance-a-binary-search-tree/
@brief
给你一棵二叉搜索树，请你返回一棵 平衡后 的二叉搜索树，新生成的树应该与原来的树有着相同的节点值。如果有多种构造方法，请你返回任意一种。
如果一棵二叉搜索树中，每个节点的两棵子树高度差不超过 1 ，我们就称这棵二叉搜索树是 平衡的 。
*/
#include <assert.h>
#include <stdio.h>
#include <algorithm>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <unordered_map>
#include <vector>
#include "traverse.h"
using namespace std;
using namespace tree_util;
class Solution
{
public:
TreeNode* balanceBST(TreeNode* root)
{
dfs(root);
return build(0, inorder_arr.size() - 1);
}
private:
TreeNode* build(int left, int right)
{
int mid = left + (right - left) / 2;
TreeNode* root = new TreeNode(inorder_arr[mid]);
if (left <= mid - 1) {
root->left = build(left, mid - 1);
}
if (mid + 1 <= right) {
root->right = build(mid + 1, right);
}
return root;
}
void dfs(TreeNode* root)
{
if (!root) return;
dfs(root->left);
inorder_arr.push_back(root->val);
dfs(root->right);
}
vector<int> inorder_arr;
};
int main() { return 0; }