BIT 1031 Binary Tree Traversals

题意就是给一个二叉树的前序遍历序列和中序遍历序列

9

1 2 4 7 3 5 8 9 6

4 7 2 1 8 5 9 3 6
对前序序列
1 2 4 7 3 5 8 9 6

1一定是该树的根节点

1在中序遍历中又在4 7 2之后

于是有



一层层递归下去就好了。。

#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
int preorder[1100];//先序
int inorder[1100];//中序
class node
{
public:
int val;
node *left;
node *right;
};
void buildTree(node *tree,int prel,int prer,int inl,int inr)
{
//此时tree里应以prel为根节点，并在i中序序列中找到prel
tree->val=preorder[prel];
tree->left=NULL;
tree->right=NULL;
int i;
for(i=inl;i<=inr;i++)
{
if(inorder[i]==preorder[prel])
{
break;
}
}
if(prel+1<=i-inl+prel&&inl<=i-1)
{
tree->left=new node;
buildTree(tree->left,prel+1,i-inl+prel,inl,i-1);
}
if(i-inl+prel+1<=prer&&i+1<=inr)
{
tree->right=new node;
buildTree(tree->right,i-inl+prel+1,prer,i+1,inr);
}
}
void postTravel(node *p)
{
if(p==NULL)
{
return;
}
postTravel(p->left);
postTravel(p->right);
printf("%d ",p->val);
}
int main()
{
int n;
while (~scanf("%d",&n))
{
for (int i = 0; i < n; i++)
{
scanf("%d",&preorder[i]);
}
for (int i = 0; i < n; i++)
{
scanf("%d",&inorder[i]);
}
node *tree;
tree=new	node;
buildTree(tree,0,n-1,0,n-1);
postTravel(tree->left);//后序遍历开始
postTravel(tree->right);
printf("%d\n",tree->val);
}
return 0;
}