面试题7：重建二叉树
对vector使用指针
#include <stdio.h>
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<int> a,b,c;
for (int i = 0; i < 3; i++)
{
a.push_back( i );
b.push_back( i + 3);
c.push_back(i + 6);
}
vector<int>* seq[3] = {&a,&b,&c};
vector<int>* curr = 0;
for (int j = 0; j < 3; j++)
{
curr = seq[j];
printf("%d\n", curr->at(0) );
printf("%d\n", curr->at(1));
printf("%d\n", curr->at(2) );
}
getchar();}
二叉树结构体定义
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
题目描述
输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。
刷第一遍
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
int lenth = pre.size();
//① 空树
if(lenth == 0 || &pre == nullptr || &vin == nullptr)
return nullptr;
Core(&pre,&pre+lenth-1,&vin,&vin+lenth-1);
}
//S第一个数，E最后一个数
TreeNode* Core(int* preS,int* preE,int* vinS,int* vinE)//bug1
{
//初试化根节点
int rootvalue = preS->val;//bug2
TreeNode* root = new TreeNode(rootvalue);
//② 只有一个根节点
if(preS == preE)
{
if(vinS == vinE && vinS->val == preS->val)
{
return root;
}
else
{
throw exception("wrong");
}
}
//求leftlength
//先找vinroot
int* vinroot = preS;
while(vinroot <= vinE && vinroot->val != rootvalue)//这里的二叉树一定不含重复数字不然就不能这样找vinroot了
++vinroot;
//③ 如果vin中没有root,报错
if(vinroot == vinE && vinroot->val != rootvalue )
throw exception("wrong");
int leftlength = vinroot - preS;
//左子树
root->left = Core(preS + 1,preS + leftlength ,vinS,vinroot - 1);
//右子树
root->right = Core(preS + leftlength + 1,preE,vinroor + 1,vinE);
return root;
}
};
1.
TreeNode* Core(int* preS,int* preE,int* vinS,int* vinE)//bug1error: cannot initialize a parameter of type 'int *' with an rvalue of type 'vector *'
Core(&pre,&pre+lenth-1,&vin,&vin+lenth-1);
修改为
TreeNode* Core(vector<int>* preS,vector<int>* preE,vector<int>* vinS,vector<int>* vinE)2.
int rootvalue = preS->val;//bug2error: no member named 'val' in 'std::vector<int, std::allocator >'
int rootvalue = preS->val;
vector<int>* 是对 vector 的指针，下一个是vector的下一位
刷第二遍
提交时间：2018-07-15 语言：C++ 运行时间： 5 ms 占用内存：612K 状态：答案正确
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
int lenth = pre.size();
//① 空树
if(lenth <= 0 || &pre == nullptr || &vin == nullptr)
return nullptr;
return Core(&pre[0],&pre[0]+lenth-1,&vin[0],&vin[0]+lenth-1);
}
//S第一个数，E最后一个数
TreeNode* Core(int* preS,int* preE,int* vinS,int* vinE)
{
//初试化根节点
int rootvalue = *preS;
TreeNode* root = new TreeNode(rootvalue);
//② 只有一个根节点
if(preS == preE)
{
if(vinS == vinE && *vinS == *preS)
{
return root;
}
else
{
throw exception();
}
}
//求leftlength
//先找vinroot
int* vinroot = vinS;//bug1:int* vinroot = preS;
while(vinroot <= vinE && *vinroot != rootvalue)//这里的二叉树一定不含重复数字不然就不能这样找vinroot了
++vinroot;
//③ 如果vin中没有root,报错
if(vinroot == vinE && *vinroot != rootvalue )
throw exception();
int leftlength = vinroot - vinS;//bug2:int leftlength = vinroot - preS;
//左子树
if(leftlength>0)//bug3忘了判断递归条件,不判断会内存超限
root->left = Core(preS + 1,preS + leftlength ,vinS,vinroot - 1);
//右子树
if(preE-preS>leftlength)//bug3忘了判断递归条件
root->right = Core(preS + leftlength + 1,preE,vinroot + 1,vinE);
return root;
}
};
优秀代码参考
public class Solution {
public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);
return root;
}
private TreeNode reConstructBinaryTree(int [] pre,int startPre,int endPre,int [] in,int startIn,int endIn) {
if(startPre>endPre||startIn>endIn)
return null;
TreeNode root=new TreeNode(pre[startPre]);
for(int i=startIn;i<=endIn;i++)
if(in[i]==pre[startPre]){
root.left=reConstructBinaryTree(pre,startPre+1,startPre+i-startIn,in,startIn,i-1);
root.right=reConstructBinaryTree(pre,i-startIn+startPre+1,endPre,in,i+1,endIn);
break;
}
return root;
}
}数组的索引与指针的思想。
Python
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回构造的TreeNode根节点
def reConstructBinaryTree(self, pre, tin):
# write code here
if not pre or not tin:
return None
root = TreeNode(pre.pop(0))
index = tin.index(root.val)
root.left = self.reConstructBinaryTree(pre, tin[:index]) //(pre[:index], tin[:index])
root.right = self.reConstructBinaryTree(pre, tin[index + 1:])//(pre[index:], tin[index+1:])
return rootpython 切片[1:5] 输出索引1~4
