1020. Tree Traversals (25)

2 3 1 5 7 6 4
1 2 3 4 5 6 7
得到root为4
左树： 2 3 1
1 2 3
右树： 5 7 6
5 6 7
递归

#include<iostream>#define MAX_Node 32#include<deque>using namespace std;int post[MAX_Node];//保存后序int in[MAX_Node];//保存中序int N;typedef struct Node{    int data;    struct Node *lchild;    struct Node *rchild;}Node,*Tree;int findN(int x,int a)//寻找root在中序某范围内的位置{    for (int t = 0;t < N;t++)        if (x == in[t+a]) return t;    return 0;}Tree findchild(int m,int n,int x,int y){    if (m > n) {return NULL; }    Node *root=(Node *)malloc(sizeof(Node));    root->data = post[n];    int mid = findN(root->data,x);    root->lchild=findchild(m, m+mid-1,x,x+mid-1);    root->rchild=findchild(m+mid, n-1, x+mid+1,y);    return root;}void bfstraverse(Node *p){    deque<Node *> que;    cout << p->data;    if (p->lchild != NULL) que.push_back(p->lchild);    if (p->rchild != NULL) que.push_back(p->rchild);    while (!que.empty())    {        if (que.front()->lchild != NULL) que.push_back(que.front()->lchild);        if (que.front()->rchild != NULL) que.push_back(que.front()->rchild);        cout << " "<<que.front()->data;        que.pop_front();    }}int main(){    Node *p;    cin >> N;    for (int t = 0;t < N;t++)        cin >> post[t];    for (int t = 0;t < N;t++)        cin >> in[t];    p=findchild(0, N - 1,0,N-1);//构建树木    bfstraverse(p);//层序遍历树    cout << endl;}