printf any exists swe ted posit bin separate tin

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

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.

Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N ($\le$

30), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first printf in a line Yes if the tree is unique, or No if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input 1:

7
1 2 3 4 6 7 5
2 6 7 4 5 3 1

Sample Output 1:

Yes
2 1 6 4 7 3 5

Sample Input 2:

4
1 2 3 4
2 4 3 1

Sample Output 2:

No
2 1 3 4

給出前序和後序遍歷，看是否能確定一棵樹，前序遍歷是根左右，後序是左右根所以通過對比可以先確定根然後左右子樹就是中間夾的那一塊，由於每一棵子樹都滿足這個條件，所以能把左右子樹給分開，可以遞歸去構建左右子樹，如果只有一棵子樹，左右子樹必有一為空，這樣這棵樹就是不確定的，至於輸出，統一當成左子樹或右子樹。 代碼：

#include <stdio.h>
int n,pre[31],post[31],in[31],c,flag = 1;
void getin(int prel,int prer,int postl,int postr) {
    if(prel > prer)return;
    int d = pre[prel ++];///根結點
    postr --;///避開根結點前移
    int t = -1;
    for(int i = postl;i <= postr;i ++) {
        if(post[i] == pre[prel]) {///確定左子樹範圍
            t = i;
            break;
        }
    }
    ///中序遍歷
    if(t != -1)getin(prel,prel + t - postl,postl,t);///左子樹
    in[c ++] = d;///根
    if(t != -1 && t != postr)getin(prel + t - postl + 1,prer,t + 1,postr);///右子樹
    else if(t != -1)flag = 0;///沒有右子樹 實際上子樹既可以是左子樹也可以是右子樹
}
int main() {
    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]);
    }
    getin(0,n - 1,0,n - 1);
    printf("%s\n",flag ? "Yes" : "No");
    for(int i = 0;i < c;i ++) {
        if(i)printf(" %d",in[i]);
        else printf("%d",in[i]);
    }
    putchar(‘\n‘);
}

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