树的遍历

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2019-08-03

20:31:37

Binary trees are defined recursively. A binary tree T is a structure defined on a finite set of nodes that either

• contains no nodes, or
• is composed of three disjoint sets of nodes:
  - a root node.
  - a binary tree called its left subtree.
  - a binary tree called its right subtree.

Your task is to write a program which perform tree walks (systematically traverse all nodes in a tree) based on the following algorithms:

1. Print the root, the left subtree and right subtree (preorder).
2. Print the left subtree, the root and right subtree (inorder).
3. Print the left subtree, right subtree and the root (postorder).

Here, the given binary tree consists of n nodes and evey node has a unique ID from 0 to n-1.

Input

The first line of the input includes an integer n, the number of nodes of the tree.

In the next n linen, the information of each node is given in the following format:

id left right

id is the node ID, left is ID of the left child and right is ID of the right child. If the node does not have the left (right) child, the left(right) is indicated by -1

Output

In the 1st line, print "Preorder", and in the 2nd line print a list of node IDs obtained by the preorder tree walk.

In the 3rd line, print "Inorder", and in the 4th line print a list of node IDs obtained by the inorder tree walk.

In the 5th line, print "Postorder", and in the 6th line print a list of node IDs obtained by the postorder tree walk.

Print a space character before each node ID.

Constraints

• 1 ≤ n ≤ 25

Sample Input 1

9
0 1 4
1 2 3
2 -1 -1
3 -1 -1
4 5 8
5 6 7
6 -1 -1
7 -1 -1
8 -1 -1

Sample Output 1

Preorder
 0 1 2 3 4 5 6 7 8
Inorder
 2 1 3 0 6 5 7 4 8
Postorder
 2 3 1 6 7 5 8 4 0

#include <bits/stdc++.h>
using namespace std;
#define NIL -1
#define MAX 100005
struct Node{
    int parent, left, right;
}t[MAX];
//前序遍历 
void preParse(int u)
{
    if(u == NIL) return ;
    printf(" %d", u);
    preParse(t[u].left);
    preParse(t[u].right);
}
//中序遍历
void inParse(int u)
{
    if(u == NIL) return;
    inParse(t[u].left);
    printf(" %d", u);
    inParse(t[u].right);    
} 
//后序遍历
void postParse(int u)
{
    if(u == NIL) return;
    postParse(t[u].left);
    postParse(t[u].right);
    printf(" %d", u);    
} 
int main()
{
    int n;
    int root;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        t[i].parent = NIL;
    }
    
    for(int i = 0; i < n; i++)
    {
        int v,l,r;
        cin >> v >> l >> r;
        t[i].parent = v;
        t[i].left = l;
        t[i].right = r;
        if(t[l].left != NIL) t[l].parent = v;
        if(t[i].right != NIL) t[r].parent = v;
    }
    
    for(int i = 0; i < n; i++)
    {
        if(t[i].parent == NIL)
        {
            root = i;
        }
    }
    
    printf("Preorder\n");
    preParse(root);
    printf("\n");
    printf("Inorder\n");
    inParse(root);
    printf("\n");
    printf("Poostorder\n");
    postParse(root);
    printf("\n");
    
    return 0;
}

树的遍历

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原文地址：https://www.cnblogs.com/Artimis-fightting/p/11296189.html