为什么搜索和后继者和前任者返回-1？
// BST.cpp : main project file.
#include "stdafx.h"
#include <cstdlib>
#include <iostream>
#define SIZE 10
using namespace std;
struct Node {
int value;
Node *left;
Node *right;
Node *parent;
};
struct BST {
Node *root;
};
void insert(int value, BST *tree) {
Node *x = tree->root;
Node *y = NULL;
Node *z = (Node *) malloc(sizeof(Node));
z->left = NULL;
z->right = NULL;
z->value = value;
// Add your code here
while (x!=NULL){
y=x;
if (z->value < x->value)
x= x->left;
else x = x->right;
}
z->parent=y;
if (y==NULL)
tree->root=z;
else if (z->value <y->value)
y->left =z;
else y->right =z;
}
Node *search(int key, Node *n) {
if (n== NULL || key == n->value)
return n;
if (key < n->value)
search(key, n->left);
else
search(key, n->right);
}
Node *min(Node *n) {
if (n == NULL || n->left == NULL)
return n;
else
return min(n->left);
}
Node *max(Node *n) {
if (n == NULL || n->right == NULL)
return n;
else
return max(n->right);
}
Node *successor(int value, Node *n) {
Node *y = NULL;
Node *x = search(value, n);
if (x == NULL)
return NULL;
if (x->right != NULL)
return min(x->right);
y = x->parent;
while (y != NULL && x == y->right) {
x = y;
y = y->parent;
}
return y;
}
Node *predecessor(int value, Node *n) {
Node *x = search(value, n);
Node *y = NULL;
if (x == NULL)
return NULL;
if (x->left != NULL)
return max(x->left);
y = x->parent;
while (y != NULL && x == y->left) {
x = y;
y = y->parent;
}
return y;
}
Node *remove(int value, BST *tree) {
Node *z = search(value, tree->root);
Node *y = NULL, *x = NULL;
if (z == NULL) return NULL;
if (z->left == NULL || z->right == NULL)
y = z;
else
y = successor(value, z);
if (y->left != NULL)
x = y->left;
else
x = y->right;
if (x != NULL)
x->parent = y->parent;
if (y->parent == NULL)
tree->root = x;
else if (y == y->parent->left)
y->parent->left = x;
else
y->parent->right = x;
if (y != z) {
int tmp = z->value;
z->value = y->value;
y->value = tmp;
}
return y;
}
// ascending sort function
void sortAsc(Node *node) {
//Add your code here
//inorder
if (node->left!=NULL)
sortAsc(node->left);
cout<<node->value<<" ";
if (node->right!=NULL)
sortAsc(node->right);
}
// descending sort function
void sortDes(Node *node) {
// Add your code here
//inorder
if (node->right!=NULL)
sortDes(node->right);
cout<<node->value<<" ";
if (node->left!=NULL)
sortDes(node->left);
}
void clear(BST *tree) {
Node *n = NULL;
while (tree->root != NULL) {
n = remove(tree->root->value, tree);
free(n);
}
}
int main() {
int A[] = {3, 5, 10, 4, 8, 9, 1, 4, 7, 6};
Node *node = NULL;
BST *tree = (BST *) malloc(sizeof(BST));
tree->root = NULL;
// build BST tree
cout << "Input data:\n\t";
for (int i=0; i<SIZE; i++) {
cout << A[i] << " "; // by the way, print it to the console
insert(A[i], tree); // You need to complete TASK 1, so that it can work
}
// sort values in ascending order
cout << "\n\nAscending order:\n\t";
sortAsc(tree->root); // You need to complete TASK 2. Otherwise you see nothing in the console
// sort values in descending order
cout << "\n\nDescending order:\n\t";
sortDes(tree->root); // TASK 2 also!
// Find minimum value
if (tree->root != NULL)
cout << "\n\nMin: " << min(tree->root)->value;
// Find maximum value
if (tree->root != NULL)
cout << "\n\nMax: " << max(tree->root)->value;
// delete 4
cout << "\n\nDelete 4 and add 2";
//free(remove(4, tree)); // You need to complete TASK 3, so that remove(int, BST *) function works properly
// we also need to release the resource!!!
// insert 2
insert(2, tree); // It belongs to TASK 1 too.
cout << "\n\nAscending order:\n\t";
sortAsc(tree->root); // TASK 2!!
// Find the successor of 5, -1 means no successor
node = search(5, tree->root);
cout << "\n\nSearch of 5 is: " << (node != NULL?node->value:-1);
// Find the successor of 5, -1 means no successor
node = successor(5, tree->root);
cout << "\n\nSuccessor of 5 is: " << (node != NULL?node->value:-1);
// Find the predecessor of 5. -1 means no predecessor
node = predecessor(5, tree->root);
cout << "\n\nPredecessor of 5 is: " << (node != NULL?node->value:-1);
cout << "\n\n";
// clear all elements
clear(tree); // delete all nodes and release resource
free(tree); // delte the tree too
system("Pause");
}
答案 0 :(得分：4)
你的递归搜索中有一个错误，你需要让所有路径都返回这样的值：
Node *search(int key, Node *n) {
if (n== NULL || key == n->value)
return n;
if (key < n->value)
return search(key, n->left);
else
return search(key, n->right);
}
除此之外，我倾向于先尝试调试自己的代码并提供有关您所发现内容的更多详细信息，而不仅仅是发布代码并询问它有什么问题。否则，你可能会得到一些真正的聪明屁股答案;）