面试题6：重建二叉树

题目链接：http://ac.jobdu.com/problem.php?pid=1385

思路：前序遍历结果的第一个数字就是根节点，找到根节点在中序遍历中
的位置，则该位置左边的即为左子树的中序遍历的结果。显然我们很容易
就可以得到左右子树的前序和中序遍历结果，我们可以使用递归来构建。

小知识：

preorder  前序序列

inorder    中序序列

postoredr 后序序列

code:

 1 #include <cstdio>

 2 using namespace std;

 3 const int MAXN = 1005;

 4 struct BinaryTreeNode

 5 {

 6     int             nValue;

 7     BinaryTreeNode* nLeft;

 8     BinaryTreeNode* nRight;

 9 };

10  

11 bool flag = true;  // 标志是否构建成功

12  

13 BinaryTreeNode* BuildTree(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder)

14 {

15     if (flag == false) return NULL;

16     int rootValue = startPreorder[0];

17     BinaryTreeNode* root = new BinaryTreeNode();

18     root->nValue = rootValue;

19     root->nLeft = root->nRight = NULL;

20  

21     // 非法数据预判

22     if (startPreorder == endPreorder)

23     {

24         if (startInorder == endInorder && *startPreorder == *startInorder) return root;

25         else

26         {

27             flag = false;

28             return NULL;

29         }

30     }

31     // 在中序遍历中找到根节点

32     int* rootInder = startInorder;

33     while (rootInder <= endInorder && *rootInder != rootValue) ++rootInder;

34     // 非法数据检查

35     if (rootInder == endInorder && *rootInder != rootValue)

36     {

37         flag = false;

38         return NULL;

39     }

40     int leftLength = rootInder - startInorder;

41     int* leftPreorderEnd = startPreorder + leftLength;

42     if (leftLength > 0)

43     {

44         // 构建左子树

45         root->nLeft = BuildTree(startPreorder + 1, leftPreorderEnd, startInorder, rootInder - 1);

46     }

47     if (leftLength < endPreorder - startPreorder)

48     {

49         // 构建右子树

50         root->nRight = BuildTree(leftPreorderEnd + 1, endPreorder, rootInder + 1, endInorder);

51     }

52     return root;

53 }

54  

55 void postorderTravel(BinaryTreeNode* root)

56 {

57     // 递归遍历

58     if (root != NULL)

59     {

60         postorderTravel(root->nLeft);

61         postorderTravel(root->nRight);

62         printf("%d ", root->nValue);

63     }

64 }

65  

66 int main()

67 {

68     int preOrder[MAXN];

69     int inOrder[MAXN];

70     int n;

71     while (scanf("%d", &n) != EOF)

72     {

73         for (int i = 0; i < n; ++i) scanf("%d", &preOrder[i]);

74         for (int i = 0; i < n; ++i) scanf("%d", &inOrder[i]);

75         flag = true;

76         BinaryTreeNode* root = BuildTree(preOrder, preOrder + n - 1, inOrder, inOrder + n - 1);

77         if (flag == false) printf("No");

78         else postorderTravel(root);

79         printf("\n");

80     }

81     return 0;

82 }