将字符串转换为数组_LeetCode 树 108.将有序数组转换为二叉搜索树

7(108) 将有序数组转换为二叉搜索树

描述

将一个按照升序排列的有序数组，转换为一棵高度平衡二叉搜索树。

本题中，一个高度平衡二叉树是指一个二叉树每个节点的左右两个子树的高度差的绝对值不超过 1。

示例

给定有序数组: [-10,-3,0,5,9],一个可能的答案是：[0,-3,9,-10,null,5]，它可以表示下面这个高度平衡二叉搜索树：0/ -3   9/   /-10  5

Description

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example

Given the sorted array: [-10,-3,0,5,9],One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:0/ -3   9/   /-10  5

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解题

二叉搜索树 BST(Binary Search Tree) 又称为二叉查找树二叉排序树二叉搜索树或者是一棵空的树或者是具有以下特征的树: 1. 若左子树非空则左子树上所有的结点key_val均小于根节点的key_val 2. 若右子树非空则右子树上所有的结点key_val均大于根节点的key_val 3. 左右子树本身也是二叉搜索树

因此对二叉搜索树进行中序遍历得到的一定是一个递增的有序序列

#include <iostream>// SortedArray to BST
#include <vector>
#include <queue>
using namespace std;
struct TreeNode{int val;TreeNode *left;TreeNode *right;TreeNode(int x):val(x),left(NULL),right(NULL) {}
};
TreeNode* levelCreateBinaryTree(const vector<int> &nums,int len,int index){//层序创建二叉树index为位置序号TreeNode *root=NULL;if(index<len&&nums[index]!=-1){root = new TreeNode(nums[index]);root->left = levelCreateBinaryTree(nums,len,2*index+1);root->right= levelCreateBinaryTree(nums,len,2*index+2);}return root;
}
void PrintMartrix(vector<vector<int>>& res){//打印二维数组for(int i=0;i<res.size();++i){cout<<"[";for(int j=0;j<res[i].size();++j){cout<<" "<<res[i][j]<<" ";}cout<<"]"<<endl;}
}
class Solution {public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> res;if(!root) return res;queue<TreeNode* > Tree;Tree.push(root);while(!Tree.empty()){vector<int> temp;int len=Tree.size();while(len--){TreeNode *pNode=Tree.front();Tree.pop();temp.push_back(pNode->val);if(pNode->left) Tree.push(pNode->left);if(pNode->right) Tree.push(pNode->right);}res.push_back(temp);}return res;}vector<int> res;vector<int> inorderTraversal(TreeNode* root) {//递归算法if(root!=NULL){inorderTraversal(root->left);res.push_back(root->val);inorderTraversal(root->right);}return res;}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* node=new TreeNode(nums[mid]);vector<int>::const_iterator first;vector<int>::const_iterator last;first=nums.begin();last=nums.begin()+mid;vector<int> v_temp(first,last);node->left=sortedArrayToBST(v_temp);if(mid==nums.size()-1) node->right=NULL;else{first=nums.begin()+mid+1;last=nums.end();vector<int> v_temp(first,last);node->right=sortedArrayToBST(v_temp);}return node;}
};
int main(){vector<int> nums={0,1,2,3,4,5};//示例 值为-1代表所在位置为空值int len=nums.size();Solution answer;TreeNode *root=answer.sortedArrayToBST(nums);vector<vector<int>> res=answer.levelOrder(root);PrintMartrix(res);//打印层序遍历的二维数组vector<int> print=answer.inorderTraversal(root);for(auto x:print) cout<<x<<" ";//打印中序遍历的序列return 0;
}