31,432 sequential traversal of a fork tree

31,432 sequential traversal of a fork tree

time limit: 1 sspace limit: 32000 KBtitle level: Silver SolvingTitle Description Description

To find a binary tree's pre-sequence traversal, middle sequence traversal and post-order traversal

Enter a description Input Description

The first line is an integer n, which indicates the number of nodes in the tree.

The next n rows are 2 integer l and r per line. The two integers of line I, Li and Ri, represent the left son number and the right son number of the node numbered I.

Output description Output Description

The output is three lines, namely, the pre-sequence traversal, the middle sequence traversal and the post-order traversal. The numbers are separated by a space.

Sample input Sample Input

5

2 3

4 5

0 0

0 0

0 0

Sample output Sample Output

1 2 4) 5 3

4 2 5) 1 3

4 5 2) 3 1

Data range and Tips Data Size & Hint

N <= 16

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#include <cstdio>inta[101][3],n;voidFrontintN) {    if(!n)return ; printf ("%d", N); Front (a[n][1]); Front (a[n][2]); }voidMiddle (intN) {    if(!n)return ; Middle (a[n][1]); printf ("%d", N); Middle (a[n][2]); }voidBehindintN) {    if(!n)return ; Behind (a[n][1]); Behind (a[n][2]); printf ("%d", N); }intMain () {scanf ("%d",&N);  for(intI=1; i<=n;i++) {a[i][0]=i; scanf ("%d%d", &a[i][1],&a[i][2]); } Front (a[1][0]); Puts (""); Middle (a[1][0]); Puts (""); Behind (a[1][0]); Puts (""); return 0;}

31,432 sequential traversal of a fork tree