二分查找树按照key值划分

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <string.h>
#include <stdio.h>
#include <unistd.h>
#include <cstdio>
#include <queue>
using namespace std;

class TreeNode{
public:
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int v):val(v),left(nullptr),right(nullptr){}
};

TreeNode* buildBST(vector<int> &data,int s,int e){
TreeNode *root = nullptr;
if(s<=e) {
int mid = (s + e) / 2;
root = new TreeNode(data[mid]);
root->left = buildBST(data,s,mid-1);
root->right = buildBST(data,mid+1,e);
}
return root;
}

bool checkBST(TreeNode *root){
if(root== nullptr){
return true;
}
int rootv = root->val;
bool isleft = true;
bool isright = true;
if(root->left){
isleft = rootv >=root->left->val? true:false;
isleft = isleft && checkBST(root->left);
}
if(root->right){
isright = rootv < root->right->val? true:false;
isright = isright && checkBST(root->right);
}
return isleft&&isright;
}

void printBST(TreeNode *root){
deque<TreeNode*> myd;
myd.push_back(root);
int num = 1;
int nextlineNumbers = 0;
while(!myd.empty()){
while(num--){
TreeNode *t = myd[0];
if(t->left){
nextlineNumbers++;
myd.push_back(t->left);
}
if(t->right){
nextlineNumbers++;
myd.push_back(t->right);
}
cout<<t->val<<" ";
myd.pop_front();
}cout<<endl;
num = nextlineNumbers;
nextlineNumbers = 0;
}//while
}

/*根据bst的性质,我们只需要对树的边界边调整即可,
* 树的边界边分为下面两种;
> //大于key
/
/
<= //小于等于key
后者是
<= //小于等于key


> //大于key
* */
//我当时卡在了BST树的性质对于这道题的认知,
//对root节点的左子树中所有的节点值都比root节点小
//对root节点的右子树中所有的节点值都比root节点大
//将边界边切开,分为左右两个树,按照性质,只需要对两棵树其中的一课继续做拆分操作即可。
vector<TreeNode*> splitByKey(TreeNode *root,int key){
vector<TreeNode*> re;
TreeNode *r1,*r2;//分别指向被拆开得新bst树的,左侧部分和右侧部分;
r1 = r2 = nullptr;
if(root== nullptr){//对于叶子节点来说,查分完后的两部分都是nullptr
re.push_back(nullptr);
re.push_back(nullptr);
return re;
}
TreeNode *curr = root;//当前访问节点,为了找到被分割的临界边,指向临界边的“上节点”
while(curr->val<=key && curr->right){//将curr节点向右下划行
if(curr->right->val > key){
break;
}
curr= curr->right;
}

while(curr->val > key && curr->left){//将curr节点向左下划行
if(curr->left->val <= key){
break;
}
curr = curr->left;
}
if(curr->val <= key){
r1 = root;
r2 = curr->right;
curr->right = nullptr;
if(r2!= nullptr){
vector<TreeNode*> t = splitByKey(r2,key);
curr->right = t[0];
r2 = t[1];
}
}else{
r1 = curr->left;
r2 = root;
curr->left = nullptr;
if(r1!= nullptr){
vector<TreeNode*> t = splitByKey(r1,key);
curr->left = t[1];
r1 = t[0];
}
}
re.push_back(r1);
re.push_back(r2);
return re;
}

int main(){
vector<int> data;
for(int i = 0;i<10;i++){
data.push_back(i);
}

TreeNode *root = buildBST(data,0,data.size()-1);
printBST(root);
cout<<"=========="<<endl;
vector<TreeNode*> re = splitByKey(root,5);
printBST(re[0]);
cout<<"check left bst: "<<checkBST(re[0])<<endl;
cout<<"==="<<endl;
printBST(re[1]);
cout<<"check right bst: "<<checkBST(re[1])<<endl;
return false;
}
原文地址:https://www.cnblogs.com/li-daphne/p/6637804.html