Sicily 3702. 二叉搜索树的遍历

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本文介绍如何使用数组实现二叉搜索树的前序和中序遍历，通过递归方法构建和遍历树结构。

二叉搜索树的遍历，老规矩，用数组来模拟（当然可以实际建一个二叉搜索树，其实也不会麻烦，只是要操作地址总感觉不太放心），然后递归进行前中序遍历。

Run Time: 0sec

Run Memory: 348KB

Code length: 1487Bytes

Submit Time: 2012-01-05 01:00:36

// Problem#: 3702
// Submission#: 1174141
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <cstring>
using namespace std;

struct Node {
    int data;
    int left, right;
} node[ 3001 ];

void preorder( int n ) {
    if ( n != 0 ) {
        cout << node[ n ].data << " ";
        preorder( node[ n ].left );
        preorder( node[ n ].right );
    }
}

void inorder( int n ) {
    if ( n != 0 ) {
        inorder( node[ n ].left );
        cout << node[ n ].data << " ";
        inorder( node[ n ].right );
    }
}

int main()
{
    int m;
    int i, j;

    while ( cin >> m && m ) {
        cin >> node[ 1 ].data;
        node[ 1 ].left = 0;
        node[ 1 ].right = 0;
        for ( i = 2; i <= m; i++ ) {
            cin >> node[ i ].data;
            node[ i ].left = 0;
            node[ i ].right = 0;

            j = 1;
            while ( j != 0 ) {
                if ( node[ i ].data < node[ j ].data ) {
                    if ( node[ j ].left == 0 ) {
                        node[ j ].left = i;
                        j = 0;
                    }
                    else
                        j = node[ j ].left;
                }
                else {
                    if ( node[ j ].right == 0 ) {
                        node[ j ].right = i;
                        j = 0;
                    }
                    else
                        j = node[ j ].right;
                }
            }
        }

        inorder( 1 );
        cout << endl;
        preorder( 1 );
        cout << endl;
    }

    return 0;

}