pat1119 Pre- and Post-order Traversals

题意:给出先序,后序,问二叉树是否唯一,输出中序。

思路:根据先序确定当前树的根,若二叉树唯一,则可以根据先序把后序划分成两个部分,第一部分为右子树,第二部分为左子树。若不唯一,怎只能划分成一个部分,作为左子树或右子树都行。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int MAX_N = 35;
int pre[MAX_N], post[MAX_N];
int n;
bool flag;
int lc[MAX_N], rc[MAX_N];

int build(int prel, int prer, int postl, int postr) {
    if (prel >= prer) return -1;
    if (prer-prel == 1) return prel;
    int p = -1;
    for (int i = postl; i < postr-1; i++) {
        if (pre[prel+1] == post[i]) {
            p = i; break;
        }
    }
    if ((p+1-postl) == 1 && (postr-1-p-1) == 0) flag = false;
    lc[prel] = build(prel+1, prel+p-postl+2, postl, p+1);
    rc[prel] = build(prel+p-postl+2, prer, p+1, postr-1);
    return prel;
}

int in[MAX_N];
int len;
void inorder(int r) {
    if (r == -1) return ;
    inorder(lc[r]);
    in[len++] = r;
    inorder(rc[r]);
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &pre[i]);
    for (int i = 0; i < n; i++) scanf("%d", &post[i]);
    memset(lc, -1, sizeof(lc));
    memset(rc, -1, sizeof(rc));
    flag = true;
    int r = build(0, n, 0, n);
    if (flag) printf("Yes\n");
    else printf("No\n");
    len = 0;
    inorder(r);
    for (int i = 0; i < len; i++) {
        if (i > 0) printf(" ");
        printf("%d", pre[in[i]]);
    }
    printf("\n");
    return 0;
}

 

American Heritage 题目描述 Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear tree in-order" and tree pre-order" notations. Your job is to create the `tree post-order" notation of a cow"s heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes. Here is a graphical representation of the tree used in the sample input and output: C / \ / \ B G / \ / A D H / \ E F The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree. The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree. The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root. ---------------------------------------------------------------------------------------------------------------------------- 题目大意: 给出一棵二叉树的中序遍历 (inorder) 和前序遍历 (preorder),求它的后序遍历 (postorder)。 输入描述 Line 1: The in-order representation of a tree. Line 2: The pre-o rder representation of that same tree. Only uppercase letter A-Z will appear in the input. You will get at least 1 and at most 26 nodes in the tree. 输出描述 A single line with the post-order representation of the tree. 样例输入 Copy to Clipboard ABEDFCHG CBADEFGH 样例输出 Copy to Clipboard AEFDBHGC c语言,代码不要有注释
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06-16
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