1127. ZigZagging on a Tree (30)

https://www.patest.cn/contests/pat-a-practise/1127

我想到一种直接暴力的方法，就是中序后序建树以后，直接用一个双向队列进行层序遍历，

//第一种思路//我的思路就是用了一个deque//从第二层开始，当遍历偶数层数时，左边队头出去，右边进来，//奇数层数时，deque右边出去，左边进来。#include <iostream>#include <cstdio>#include <deque>using namespace std;const int maxn = 35;int in[maxn];int post[maxn];struct node {int data;node *lchild;node *rchild;};//inOrder和postOrder 创建 树/*    inOrder:12 11 20 17 1 15 8 5    postOrder:12 20 17 11 15 8 5 1*/// use the inOrder and postOrder create the binary tree//node *Create(int postL, int postR, int inL, int inR) {if ( postL > postR ) return NULL;node *root = new node();root->data = post[postR];int index;for ( index = inL; index <= inR; index++ ) {if ( in[index] == post[postR] ) break;}int numLeft = index - inL;root->lchild = Create(postL, postL+numLeft-1, inL, index-1);root->rchild = Create(postL+numLeft, postR-1, index+1, inR);return root;}// 1 11 5 12 17 8 20 15void SpecialLevelOrder(node *root) {    deque<node*> dq;    node *T;    if ( !root ) return;    dq.push_back(root);//1    bool cnt = false;//来判断偶数和奇数    int n = 0;    //因为末尾不能有空格    T = dq.front();    dq.pop_front();    printf("%d",T->data);    if ( T->lchild ) dq.push_back(T->lchild);    if ( T->rchild ) dq.push_back(T->rchild);    cnt = true;    while ( !dq.empty() ) {        n = dq.size();        if ( cnt ) {            for ( int i = 0; i < n; i++) {                T = dq.front();                printf(" %d",T->data);                if ( T->lchild ) dq.push_back(T->lchild);                if ( T->rchild ) dq.push_back(T->rchild);                dq.pop_front();            }            cnt = false;        }        else { //cnt%2 == 0            for ( int i = 0; i < n; i++ ) {                T = dq.back();                printf(" %d",T->data);                if ( T->rchild ) dq.push_front(T->rchild);                if ( T->lchild ) dq.push_front(T->lchild);                dq.pop_back();            }            cnt = true;        }    }}int main() {int n;cin >> n;for ( int i = 0 ; i < n; i++ ) {cin >> in[i];}for ( int i = 0; i < n; i++ ) {cin >> post[i];}node *root;    root = Create(0,n-1,0,n-1);    SpecialLevelOrder(root);return 0;}

然后还有一种方法就是利用递归建树的特点，用vector记录下每一层的节点value

#include <iostream>#include <vector>using namespace std;const int maxn = 35;int post[maxn];int in[maxn];vector<int> level[maxn];//每个vector用于存储每一层的个数int len = 0;int maxx = 0;//记录层数struct node{int v;int l,r;};void SpecialPre(int root, int start, int ed, int index){if( start > ed ) return;int i = start;while( post[root] != in[i] && i < ed) i++;    maxx = max(maxx, index);level[index].push_back(post[root]);    //root-end == 7-7 == 0 +SpecialPre(root-ed+i-1, start, i-1, index+1);SpecialPre(root-1, i+1, ed, index+1);}int main(){int n;cin >> n;for( int i = 0; i < n; i++ )        cin >> in[i];for( int i = 0; i < n; i++ )        cin >> post[i];//post->root, in->start, in->end, indexSpecialPre(n-1,0,n-1,1);int cnt = n;int tt = 0;for( int i = 1; i <= maxx; i++ ) {if( i%2 == 0 ) {            for(int j = 0; j < level[i].size(); j++ ) {                if( tt == 0 ) cout << level[i][j];                else cout << " " << level[i][j];  tt++;  }}else {for(int j = level[i].size()-1; j >= 0; j--) {                if( tt == 0 ) cout << level[i][j];                else cout << " " << level[i][j];                tt++;            }}}    return 0;}