从前序遍历序列恢复BST

相关问题：已知二叉树的前序遍历序列和中序遍历序列，要求重新恢复该二叉树。

给你一个二叉搜索树的前序遍历序列，恢复此二叉搜索树

文章末尾给出了一个复杂度是O(n)的算法。复杂度是O(n^2)的代码如下：

#include <stdio.h>#include <stdlib.h>#include <vector>using namespace std; struct Node{    int key;   Node* left;  Node* right;    Node(int k, Node* l, Node* r): key(k), left(l), right(r){};};Node* construct(vector<int> pre){if(pre.size()){int data = pre[0];vector<int> pre_l, pre_r;if(pre.size()>1)for(int i=1; i<pre.size(); i++)if(pre[i]<pre[0])pre_l.push_back(pre[i]);elsepre_r.push_back(pre[i]);Node* root = new Node(data, NULL, NULL);root->left = construct(pre_l);root->right = construct(pre_r);return root;}elsereturn NULL;}void print(Node* root){if(root){print(root->left);   printf("%d|", root->key);print(root->right);}} int main(){int pre[] = {10, 5, 1, 7, 40, 50};vector<int> pre_vec = vector<int>(pre, pre+6);Node* root = construct(pre_vec);print(root);}

复杂度是O(n)的算法。思路：

 The trick is to set a range {min.. max} for every node. Initialize the range as {INT_MIN .. INT_MAX}.The first node will definitely be in range, so create root node. Toconstruct the left subtree, set the range as {INT_MIN …root->data}.If a values is in the range {INT_MIN .. root->data}, the values ispart part of left subtree. To construct the right subtree, set therange as {root->data..max .. INT_MAX}.

以下是网上某人的代码。我看不明白为何constructTreeUtil 的参数列表里 int key 的作用。

// A recursive function to construct BST from pre[]. preIndex is used// to keep track of index in pre[].Node* constructTreeUtil( int pre[], int* preIndex, int key,                                int min, int max, int size ){    // Base case    if( *preIndex >= size )        return NULL;      Node* root = NULL;      // If current element of pre[] is in range, then    // only it is part of current subtree    if( key > min && key < max ) {         root = new Node (key, NULL, NULL );        *preIndex = *preIndex + 1;                     if (*preIndex < size) {            root->left = constructTreeUtil( pre, preIndex, pre[*preIndex],                                        min, key, size );            root->right = constructTreeUtil( pre, preIndex, pre[*preIndex],                                         key, max, size );        }       }       return root;} // The function to construct BST from given preorder traversal.Node *constructTree (int pre[], int size){    int preIndex = 0;    return constructTreeUtil ( pre, &preIndex, pre[0], INT_MIN, INT_MAX, size );}int main(){        int pre[] = {10, 5, 1, 7, 40, 50};        Node* root = constructTree(pre, 6);        print(root);}

我把constructTreeUtil 的参数 int key 去掉之后，新的代码代码如下（以下代码经过测试）：

#include<stdio.h>#include<stdlib.h>#include <iostream> struct Node  {      int key;   Node* left;  Node* right;      Node(int k, Node* l, Node* r): key(k), left(l), right(r){};  };    // A recursive function to construct BST from pre[]. preIndex is used// to keep track of index in pre[].Node* constructTreeUtil( int pre[], int* preIndex,                                 int min, int max, int size ){    // Base case    if( *preIndex >= size )        return NULL;      Node* root = NULL;  int key = pre[*preIndex];    // If current element of pre[] is in range, then    // only it is part of current subtree    if( key > min && key < max ) {         root = new Node (key, NULL, NULL );        *preIndex = *preIndex + 1;                     if (*preIndex < size) {            root->left = constructTreeUtil( pre, preIndex,                                        min, key, size );            root->right = constructTreeUtil( pre, preIndex,                                         key, max, size );        }       }       return root;} // The function to construct BST from given preorder traversal.Node *constructTree (int pre[], int size){    int preIndex = 0;    return constructTreeUtil ( pre, &preIndex, INT_MIN, INT_MAX, size );}void print(Node* root){      if(root){          print(root->left);   printf("%d|", root->key);          print(root->right);      }  }int main(){        int pre[] = {10, 8, 7, 6, 9, 100, 20, 30, 40, 35, 110};        Node* root = constructTree(pre, 11);        print(root);int x; std::cin>>x;}