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:

#include <cstdio>
using namespace std;
const int MAXN = 1005;
struct BinaryTreeNode
{
    int             nValue;
    BinaryTreeNode* nLeft;
    BinaryTreeNode* nRight;
};
 
bool flag = true;  // 标志是否构建成功
 
BinaryTreeNode* BuildTree(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder)
{
    if (flag == false) return NULL;
    int rootValue = startPreorder[0];
    BinaryTreeNode* root = new BinaryTreeNode();
    root->nValue = rootValue;
    root->nLeft = root->nRight = NULL;
 
    // 非法数据预判
    if (startPreorder == endPreorder)
    {
        if (startInorder == endInorder && *startPreorder == *startInorder) return root;
        else
        {
            flag = false;
            return NULL;
        }
    }
    // 在中序遍历中找到根节点
    int* rootInder = startInorder;
    while (rootInder <= endInorder && *rootInder != rootValue) ++rootInder;
    // 非法数据检查
    if (rootInder == endInorder && *rootInder != rootValue)
    {
        flag = false;
        return NULL;
    }
    int leftLength = rootInder - startInorder;
    int* leftPreorderEnd = startPreorder + leftLength;
    if (leftLength > 0)
    {
        // 构建左子树
        root->nLeft = BuildTree(startPreorder + 1, leftPreorderEnd, startInorder, rootInder - 1);
    }
    if (leftLength < endPreorder - startPreorder)
    {
        // 构建右子树
        root->nRight = BuildTree(leftPreorderEnd + 1, endPreorder, rootInder + 1, endInorder);
    }
    return root;
}
 
void postorderTravel(BinaryTreeNode* root)
{
    // 递归遍历
    if (root != NULL)
    {
        postorderTravel(root->nLeft);
        postorderTravel(root->nRight);
        printf("%d ", root->nValue);
    }
}
 
int main()
{
    int preOrder[MAXN];
    int inOrder[MAXN];
    int n;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 0; i < n; ++i) scanf("%d", &preOrder[i]);
        for (int i = 0; i < n; ++i) scanf("%d", &inOrder[i]);
        flag = true;
        BinaryTreeNode* root = BuildTree(preOrder, preOrder + n - 1, inOrder, inOrder + n - 1);
        if (flag == false) printf("No");
        else postorderTravel(root);
        printf("
");
    }
    return 0;
}