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。
时间限制:1秒 空间限制:32768K 热度指数:335630
思想:利用递归
#include<iostream>
#include<vector>
#include<stack>
#include<stdio.h>
#include<stdlib.h>
#include<string>
using namespace std;
//Definetion 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){
if(vin.size()==0)//递归回到头条件
//if(pre.size()==0)//if(pre.empty())//if(in.empty())
return NULL;
vector<int> in_left,in_right,pre_left,pre_right;
int cur=pre[0];//前序遍历的第一个结点就是根结点
TreeNode* root=new TreeNode(cur);
int p=0;
for(;p<vin.size();p++){
if(vin[p]==cur) //p记录根节点在中序遍历中的位置
break;
}
for(int i=0;i<vin.size();i++){
if(i<p){
in_left.push_back(vin[i]);
pre_left.push_back(pre[i+1]);
}
else if(i>p){
in_right.push_back(vin[i]);
pre_right.push_back(pre[i]);
}
}
root->left=reConstructBinaryTree(pre_left,in_left);
root->right=reConstructBinaryTree(pre_right,in_right);
return root;
}
void preOrder(TreeNode* &T){
if(T==NULL) return;
else{
cout<<T->val<<" ";
preOrder(T->left);
preOrder(T->right);
}
}
};
int main(){
solution s;
TreeNode* T;
int a[8]={1,2,4,7,3,5,6,8};
vector<int> pre(&a[0],&a[8]);
int b[8]={4,7,2,1,5,3,8,6};
vector<int> in(&b[0],&b[8]);
T=s.reConstructBinaryTree(pre,in);
cout<<"创建成功!"<<endl;
cout<<"前序遍历结果:"<<endl;
s.preOrder(T);
return 0;
}
参考博客:https://blog.csdn.net/u013686654/article/details/73729288