04-树6 Complete Binary Search Tree (30分)
04-树6 Complete Binary Search Tree (30分)
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
- Both the left and right subtrees must also be binary search trees.
A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.
Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input:
10 1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
提交代码:
#include <stdio.h> #include <stdlib.h> #include <math.h> typedef int ElemType; int GetLeftLength(int N){ if (N == 1) { return 0; } int H = floor(log10(N + 1) / log10(2)); int maxX = pow(2, H - 1); int X = N - maxX * 2 + 1; X = X < maxX ? X : maxX; int L = maxX + X - 1; return L; } void Solve(int ALeft, int ARight, int TRoot, int* A, int* T){ int n = ARight - ALeft + 1; if(n==0){ return; } int L = GetLeftLength(n); T[TRoot] = A[ALeft + L]; int LeftTRoot = TRoot * 2 + 1; int RightTRoot = LeftTRoot + 1; Solve(ALeft, ALeft + L-1, LeftTRoot, A, T); Solve(ALeft + L + 1, ARight, RightTRoot, A, T); } int compare(const void*a, const void*b){ return *(int*)a-*(int*)b; } int main(){ int N; scanf("%d", &N); int *A = (int*)malloc(sizeof(ElemType)*N); int *T = (int*)malloc(sizeof(ElemType)*N); for(int i = 0; i < N; ++i){ scanf("%d", A+i); } qsort(A, N, sizeof(int), compare); Solve(0,N-1,0,A,T); if(N <= 0) return 0; printf("%d", T[0]); for(int i = 1; i < N; ++i){ printf(" %d", T[i]); } return 0; }
提测结果:
2021-08-02更新:
对完全二叉树进行中序遍历时赋值就可以了;比之前的方法简单多了!
/* 解题思路: 中序遍历序列转层序遍历序列。 */ #include <stdio.h> int Index = 0; int N = 0; int seq[1010]; int CBST[1010]; //中序遍历 void InOrder(int root, int n){ if(root > n){ return ; } InOrder(root*2, n); CBST[root] = seq[Index++]; InOrder(root*2+1, n); } //层序遍历 void LevelOrderTraversal(){ if(N != 0){ printf("%d", CBST[1]); } for(int i = 2; i <= N; ++i){ printf(" %d", CBST[i]); } } int compare(const void*a, const void*b){ return *(int*)a-*(int*)b; } int main(){ scanf("%d", &N); for(int i = 0; i < N; ++i){ scanf("%d", seq + i); } //排序 qsort(seq, N, sizeof(int), compare); InOrder(1, N); LevelOrderTraversal(); return 0; }
提交结果: