算法导论15.5-1的答案及C++实现

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①设计的伪代码CONSTRUCT-OPTIMAL-BST(root)如下

CONSTRUCT-OPTIMAL-BST(root,i,j,flag)    if i<=j        if flag == 1            print k root[i,j]为根        elseif flag == 0             print k root[i,j]为k j+1 的左孩子        else            print k root[i,j]为k i-1 的右孩子        CONSTRUCT-OPTIMAL-BST(root,i,root[i,j]-1,0)        CONSTRUCT-OPTIMAL-BST(root,root[i,j]+1,j,2)    elseif j=i-1        if flag == 0            print d j 为k i+1的左孩子        elseif flag == 2            print d j 为k i-1的右孩子

②C++实现：

#include<iostream>using namespace std;int root[6][6] = { 0, 0, 0, 0, 0, 0,0, 1, 1, 2, 2, 2,0, 0, 2, 2, 2, 4,0, 0, 0, 3, 4, 5,0, 0, 0, 0, 4, 5,0, 0, 0, 0, 0, 5,};void CONSTRUCT_OPTIMAL_BST(int i, int j, int flag){if (i <= j){if (flag == 1){cout << "k" << root[i][j] << "为根" << endl;}else if (flag == 0){cout << "k" << root[i][j] << "为k" << j + 1 << "的左孩子" << endl;}else{cout << "k" << root[i][j] << "为k" << i - 1 << "的右孩子" << endl;}CONSTRUCT_OPTIMAL_BST(i, root[i][j] - 1, 0);CONSTRUCT_OPTIMAL_BST(root[i][j] + 1, j, 2);}else if (j == (i - 1)){if (flag == 0){cout << "d" << j << "为k" << j + 1 << "的左孩子" << endl;}else if (flag == 2){cout << "d" << j << "为k" << i - 1 << "的右孩子" << endl;}}}void main(){CONSTRUCT_OPTIMAL_BST(1, 5, 1);}

输出结果为

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