5-7 Complete Binary Search Tree

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根据完全二叉搜索树的性质，将输入的数据进行从小到大排序排序，计算出左子树的规模，然后求出根节点的位置，然后把根节点放入数组，然后递归左右儿子节点，

注意求左子树的个数函数里传入的n是会随着递归变化而变化的，之前未注意到这一点，所以一直部分正确

#include <stdio.h>#include <stdlib.h>#include<algorithm>#include<math.h>using namespace std;void Solve(int ALeft, int ARight, int TRoot, int A[], int T[]);int Get_Left_Nodes(int n);int Min(int a, int b);int main(int argc, char const argv[]){    // freopen("test.txt", "r", stdin);    int N, tmp;    scanf("%d", &N);    int A[N], T[N];    for (int i = 0; i < N; ++i){        scanf("%d", &tmp);        A[i] = tmp;    }    sort(A,A+N);    int ALeft = 0, ARight = N -1, TRoot = 0;    Solve(ALeft, ARight, TRoot, A, T);    for (int i = 0; i < N; ++i)    {        if(i == 0)        {            printf("%d", T[i]);        }        else            printf(" %d", T[i]);    }    return 0;}void Solve(int ALeft, int ARight, int TRoot, int A[], int T[]){    int n;    n = ARight - ALeft + 1; //结点数n    if(n == 0)        return;    int L, LeftTRoot, RightTRoot;    L = Get_Left_Nodes(n); //计算出n个结点的完全二叉树的左子树的结点个数    T[TRoot] = A[ALeft + L];    LeftTRoot = TRoot *2 + 1;    RightTRoot = LeftTRoot + 1;    Solve(ALeft, ALeft + L - 1, LeftTRoot, A, T);    Solve(ALeft + L + 1, ARight, RightTRoot, A, T);//Aleft不要忘了+1}int Get_Left_Nodes(int n){    int H = 0, tmp = 1, X, L;//X为左子树最下一层的结点数    int N = n;    while(N > 1)    {        N /= 2;        H++;//树的高度    }    for (int i = 0; i < H - 1; ++i)    {        tmp *= 2;    }    X = n - 2*tmp + 1;    X = Min( X, tmp );    L = tmp - 1 + X;    return L;}int Min(int a, int b){    return (a < b) ? a : b;}

看了别人的之后，发现一棵二叉排序树的中序遍历序列是递增有序的
#include<stdio.h>#include<stdlib.h>#include<iostream>#include<queue>#include<algorithm>#include<map>#include<vector>#include<cstring>#include<math.h>#include<stack>using namespace std;int node[1005];int tree[1005];int pos,n;void build(int root){    if(root > n) return ;    int lson=root*2;    int rson=root*2+1;    build(lson);    tree[root]=node[pos++];    build(rson);}int main(){    cin>>n;    for(int i=0;i<n;i++)    {        cin>>node[i];    }    sort(node,node+n);    pos=0;    build(1);    for(int i=1;i<=n;i++)    {        if(i!=1)            cout<<" ";        cout<<tree[i];    }}