二叉树建树

给出前序和中序建树：

Node* build (int n, int* pre, int* in) {    Node* node = new Node;    int i = 0;    if (n <= 0)        return NULL;    while (in[i] != pre[0])        i++;    node -> val = in[i];    node -> left = build(i, pre + 1, in);    node -> right = build(n - i - 1, pre + i + 1, in + i + 1);    return node;}

给出中序和后序建树：

Node* build (int n, int* in, int* pos) {    Node* node = new Node;    int i = n - 1;    if (n <= 0)        return NULL;    while (in[i] != pos[n - 1])        i--;    node -> val = ino[i];    node -> left = build(i, in, pos);    node -> right = build(n - i - 1, in + i + 1, pos + i);    return node;}

UVA-548 给你一棵树的中序和后序遍历，求从根到叶子组成的路径中数字和最小的那条,输出最小路径的叶子。
思路：在重建完二叉树后，dfs遍历找到最小路径和叶子

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;int in[10010],pos[10010];int min1,ans;struct tree{    int data;    tree *l,*r;};tree *root;tree* build(int n, int* in, int* pos){    tree* node = new tree;    int i=n-1;    if (n<=0)        return NULL;    while (in[i]!=pos[n-1])        i--;    node->data=in[i];    node->l=build(i,in, pos);    node->r=build(n-i-1,in+i+1,pos+i);    return node;}void dfs(tree *tr,int sum){    if(tr==NULL)    {        return;    }    sum=sum+tr->data;    if(tr->l==NULL&&tr->r==NULL)    {        if(min1>sum)        {            ans=tr->data;            min1=sum;        }    }    else    {        if(tr->l!=NULL)        {            dfs(tr->l,sum);        }        if(tr->r!=NULL)        {            dfs(tr->r,sum);        }    }}int main(){    char ch;    while(scanf("%d%c",&in[0],&ch)!=EOF)    {        int n=1;        for(int i=1;ch!='\n';i++)        {            scanf("%d%c",&in[i],&ch);            n++;        }        ch = ' ';        for(int i=0;ch!='\n';i++)        {            scanf("%d%c",&pos[i],&ch);        }        min1=1000001;        root=build(n,in,pos);        dfs(root,0);        printf("%d\n",ans);    }    return 0;}

杭电1710
根据前序遍历和中序遍历，输出后序遍历

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>using namespace std;const int N=1010;int pre[N],in[N];struct tree{    int data;    tree *l,*r;};tree *root;tree *Creattree(int *pre,int *in,int n){    tree *tem;    for(int i=0;i<n;i++)    {        if(pre[0]==in[i])        {            tem=(tree*)malloc(sizeof(tree));            tem->data=in[i];            tem->l=Creattree(pre+1,in,i);            tem->r=Creattree(pre+i+1,in+i+1,n-(i+1));            return tem;        }    }    return NULL;}void Printfpos(tree *rt){    if(rt!=NULL)    {        Printfpos(rt->l);        Printfpos(rt->r);        if(rt==root)        {            printf("%d\n",rt->data);        }        else        {            printf("%d ",rt->data);        }    }}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=0;i<n;i++)        {            scanf("%d",&pre[i]);        }        for(int i=0;i<n;i++)        {            scanf("%d",&in[i]);        }        root=Creattree(pre,in,n);        Printfpos(root);    }    return 0;}