从给定的Inorder和Preorder遍历中打印Postorder遍历
给定有序的树程序的前序和后序，必须找到后尾遍历并打印
Input:
Inorder traversal in[] = {4, 2, 5, 1, 3, 6}
Preorder traversal pre[] = {1, 2, 4, 5, 3, 6}
Output:
Postorder traversal post[] = {4, 5, 2, 6, 3, 1}算法
START
Step 1 -> declare function as find_value(int p, int in_order[], int n)
Loop For i=0 and i<n and ++i
IF in_order[i]==p
Return i
End IF
End
Step 2 -> declare function as postorder(int pre_order[], int in_order[], int n)
Declare int variable as root = find_value(pre_order[0], in_order, n)
IF root!=0
Call postorder(pre_order+1, in_order, root)
End
IF root !=n-1
Call postorder(pre_order+root+1, in_order+root+1,n-root-1)
End
Print pre_order[0]
End
Step 3 -> goto main() Declare int pre_order[] = {1, 2, 4, 5, 3, 6}
Declare int in_order[] = {4, 2, 5, 1, 3, 6}
Declare int size = sizeof(pre_order)/sizeof(pre_order[0])
Call postorder(pre_order, in_order, size)
STOP示例
#include <stdio.h>
int find_value(int p, int in_order[], int n) {
for (int i = 0; i < n; ++i) {
if (in_order[i] == p) {
return i;
}
}
return -1;
}
int postorder(int pre_order[], int in_order[], int n) {
int root = find_value(pre_order[0], in_order, n);
if(root !=0 )
postorder(pre_order+1, in_order, root);
if (root != n-1)
postorder(pre_order+root+1, in_order+root+1, n-root-1);
printf("%d ", pre_order[0]);
}
int main(int argc, char const *argv[]) {
int pre_order[] = {1, 2, 4, 5, 3, 6};
int in_order[] = {4, 2, 5, 1, 3, 6};
int size = sizeof(pre_order)/sizeof(pre_order[0]);
postorder(pre_order, in_order, size);
return 0;
}输出结果
如果我们运行上面的程序，那么它将生成以下输出
4 5 2 6 3 1