1119. Pre- and Post-order Traversals (30)

来源：互联网发布：怎么测算淘宝单坑产出编辑：程序博客网时间：2024/05/22 13:19

有个坑，数字不是按1.2.3这样排序的

#include<iostream>#include<vector>#include<algorithm>using namespace std;struct node {    int data;    node *l, *r;    node() { l = r = NULL; }};vector<node> all;vector<int> pre;vector<int> post;vector<int> dat;int N,flag=0;void change(int a, int b, int c, int d){    if (a >= b||c>=d) return;    if (pre[a + 1] == post[d - 1])    {        flag = 1;        all[pre[a]].r = &all[pre[a + 1]];        change(a + 1, b, c, d - 1);    }    else     {        all[pre[a]].l = &all[pre[a + 1]];        all[pre[a]].r = &all[post[d - 1]];        int pos_pre = find(pre.begin()+1, pre.end(), post[d - 1])-pre.begin();        int pos_post= find(post.begin()+1, post.end(), pre[a+1])-post.begin();        change(a+1, pos_pre - 1, c, pos_post);        change(pos_pre, b, pos_post+1, d-1);    }}int cnt = 0;void InOrder(node *root){    if (root == NULL) return;    InOrder(root->l);    if (!cnt++) printf("%d", root->data);    else printf(" %d", root->data);    InOrder(root->r);}int find_index(int x){    for (int t = 1;t <= N;t++)        if (x == all[t].data) return t;}int main(){    cin >> N;    all.resize(N+1);    pre.resize(N+1);    post.resize(N+1);    dat.resize(N + 1);    for (int t = 1;t <= N;t++)    {        cin >> all[t].data;        pre[t]=t;    }    for (int t = 1;t <= N;t++) {        int tem;        cin >> tem;        post[t] = find_index(tem);    }    change(1, N, 1, N);    if (flag == 1)        cout << "No" << endl;    else cout << "Yes" << endl;    node *root = &all[pre[1]];    InOrder(root);    cout << endl;}

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