【编程题目】输入一颗二元树,从上往下按层打印树的每个结点,同一层中按照从左往右的顺序打印

第 16 题(树):
题目(微软):
输入一颗二元树,从上往下按层打印树的每个结点,同一层中按照从左往右的顺序打印。
例如输入
8
/ \
6 10
/ \ / \
5 7 9 11
输出 8 6 10 5 7 9 11。

 

思路:之前做USACO和学算法的时候经常用的 广度优先搜索

  1 /*
  2 第 16 题(树):
  3 题目(微软):
  4 输入一颗二元树,从上往下按层打印树的每个结点,同一层中按照从左往右的顺序打印。 
  5 例如输入
  6    8
  7  /   \
  8 6    10
  9 / \ /  \
 10 5 7 9 11
 11 输出 8 6 10 5 7 9 11。
 12 start time = 19: 10
 13 end time = 19:36
 14 */
 15 
 16 
 17 //思路 广度优先遍历
 18 #include <iostream>
 19 #include <queue>
 20 #include <stdlib.h>
 21 using namespace std;
 22 
 23 typedef struct BSTreeNode
 24 {
 25 int m_nValue; // value of node
 26 BSTreeNode *m_pLeft; // left child of node
 27 BSTreeNode *m_pRight; // right child of node
 28 }BSTreeNode;
 29 
 30 int addBSTreeNode(BSTreeNode * &T, int data)  //把data加入的以T为根的树中
 31 {
 32     if(T == NULL) //根节点单独处理
 33     {
 34         T = (BSTreeNode *)malloc(sizeof(BSTreeNode));
 35         T->m_nValue = data;
 36         T->m_pLeft = NULL;
 37         T->m_pRight = NULL;
 38     }
 39     else
 40     {
 41         BSTreeNode * x = T;
 42         BSTreeNode * px = NULL;
 43         while(x != NULL)
 44         {
 45             if(data >= x->m_nValue)
 46             {
 47                 px = x;
 48                 x = x->m_pRight;
 49             }
 50             else
 51             {
 52                 px = x;
 53                 x = x->m_pLeft;
 54             }
 55         }
 56 
 57         if(data >= px->m_nValue)
 58         {
 59             px->m_pRight = (BSTreeNode *)malloc(sizeof(BSTreeNode));
 60             px->m_pRight->m_nValue = data;
 61             px->m_pRight->m_pLeft = NULL;
 62             px->m_pRight->m_pRight = NULL;
 63         }
 64         else
 65         {
 66             px->m_pLeft = (BSTreeNode *)malloc(sizeof(BSTreeNode));
 67             px->m_pLeft->m_nValue = data;
 68             px->m_pLeft->m_pLeft = NULL;
 69             px->m_pLeft->m_pRight = NULL;
 70         }
 71     }
 72     return 1;
 73 }
 74 
 75 int BFSPrintTree(BSTreeNode * T)
 76 {
 77     queue<BSTreeNode *> Q;
 78     Q.push(T);
 79 
 80     while(!Q.empty())
 81     {
 82         BSTreeNode * x = Q.front();
 83         Q.pop();
 84         printf("%d ", x->m_nValue);
 85         if(x->m_pLeft != NULL)
 86         {
 87             Q.push(x->m_pLeft);
 88         }
 89         if(x->m_pRight != NULL)
 90         {
 91             Q.push(x->m_pRight);
 92         }
 93         
 94     }
 95     return 1;
 96 }
 97 
 98 int main()
 99 {
100     BSTreeNode * T = NULL;
101     addBSTreeNode(T, 8);
102     addBSTreeNode(T, 6);
103     addBSTreeNode(T, 10);
104     addBSTreeNode(T, 5);
105     addBSTreeNode(T, 7);
106     addBSTreeNode(T, 9);
107     addBSTreeNode(T, 11);
108 
109     BFSPrintTree(T);
110     return 0;
111 
112 }
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posted @ 2014-08-04 19:41  匡子语  阅读(456)  评论(0编辑  收藏  举报