7(108) 將有序數(shù)組轉(zhuǎn)換為二叉搜索樹(shù)

描述

將一個(gè)按照升序排列的有序數(shù)組，轉(zhuǎn)換為一棵高度平衡二叉搜索樹(shù)。

本題中，一個(gè)高度平衡二叉樹(shù)是指一個(gè)二叉樹(shù)每個(gè)節(jié)點(diǎn) 的左右兩個(gè)子樹(shù)的高度差的絕對(duì)值不超過(guò) 1。

示例

給定有序數(shù)組: [-10,-3,0,5,9],一個(gè)可能的答案是：[0,-3,9,-10,null,5]，它可以表示下面這個(gè)高度平衡二叉搜索樹(shù)：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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解題

二叉搜索樹(shù) BST(Binary Search Tree) 又稱為二叉查找樹(shù) 二叉排序樹(shù) 二叉搜索樹(shù)或者是一棵空的樹(shù) 或者是具有以下特征的樹(shù): 1. 若左子樹(shù)非空 則左子樹(shù)上所有的結(jié)點(diǎn)key_val均小于根節(jié)點(diǎn)的key_val 2. 若右子樹(shù)非空 則右子樹(shù)上所有的結(jié)點(diǎn)key_val均大于根節(jié)點(diǎn)的key_val 3. 左右子樹(shù)本身也是二叉搜索樹(shù)

因此 對(duì)二叉搜索樹(shù)進(jìn)行中序遍歷 得到的一定是一個(gè)遞增的有序序列

#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){//層序創(chuàng)建二叉樹(shù)index為位置序號(hào)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){//打印二維數(shù)組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) {//遞歸算法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 值為-1代表所在位置為空值int len=nums.size();Solution answer;TreeNode *root=answer.sortedArrayToBST(nums);vector<vector<int>> res=answer.levelOrder(root);PrintMartrix(res);//打印層序遍歷的二維數(shù)組vector<int> print=answer.inorderTraversal(root);for(auto x:print) cout<<x<<" ";//打印中序遍歷的序列return 0; }

總結(jié)

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