Binary tree algorithm: sequence and order derivation (array recursive implementation)

Reconstruction of binary tree with root sequence and posterior root sequence

Describe

We know how to travel around a binary tree in three depth priorities to get the root sequence, the previous root sequence, and the posterior root sequence. Conversely, if a binary tree is given the root sequence and the post root sequence, or given the root sequence and the previous root sequence, a one or two-fork tree can be reconstructed. The root sequence and the post root sequence of a binary tree are required to reconstruct the binary tree in memory, and finally the root sequence of the binary tree is output.

Each node of a binary tree is uniquely identified with a different integer, and the following two-tree

The root sequence is 9 5 32 67

Post-root sequence 9 32 67 5

Previous root sequence 5 9 67 32

First reading into a number n means that there are n elements in both the middle order and the post sequence.

Then enter the sequence of the middle sequence in the input sequence.

The output sequence of first order.

Input:

4

9 5 32 67

9 32 67 5

Output:

5 9 67 32

#include <stdio.h> #include <string.h>void build (int len, int *s1, int *s2, int *s) {    int p;    int i;    if (len<=0)        return;    else{for        (i=0; i<len; i++) {            if (s1[i]==s2[len-1]) {                p = i;            }        }        Build (p, S1, S2, s+1);        Build (Len-p-1, s1+p+1, S2+p, s+p+1);        S[0] = s2[len-1];}    } int main () {    int i;    int s1[1000], s2[1000], s3[1000];    int n;    while (scanf ("%d", &n)!=eof)    {for        (i=0; i<n; i++) {            scanf ("%d", &s1[i]);        }        for (i=0; i<n; i++) {            scanf ("%d", &s2[i]);        }        Build (n, s1, S2, S3);        for (i=0; i<n; i++) {            printf ("%d%c", S3[i], i==n-1? ' \ n ': ');        }    }    return 0;}

Binary tree algorithm: sequence and order derivation (array recursive implementation)