树的重建

任给一棵二叉树合法的两种遍历结果，就可以将它重建出来。事实上，不需重建这棵树，而进行隐式的遍历。可以尝试下面的这道题目：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=104&page=show_problem&problem=489

对于编程之美3.9节的这个问题及其扩展问题，简单的实现代码如下：

对扩展问题1，显然是不唯一的，考虑所有节点都标注a的极端情况即知。实现方面，需要把下面的find查找修改为直到没有匹配为止，然后分别尝试重建。

template <class T>
struct Node {
    T v;
    Node *left, *right;
};

template<class It>
bool buildable(It preFirst, It preLast, It inFirst, It inLast) {
    if (preFirst == preLast && inFirst == inLast) return true;  // both empty
    if (preFirst == preLast || inFirst == inLast) return false; // one empty

    It inMid = find(inFirst, inLast, *preFirst);
    if (inMid == inLast) return false;
    It preMid = advance(preFirst, distance(inFirst, inMid) + 1);

    if (!buildable(++preFirst, preMid, inFirst, inMid)) return false;
    return buildable(preMid, preLast, ++inMid, inLast);
}

template <class It>
Node<typename iterator_traits<It>::value_type>*
buildTree(It preFirst, It preLast, It inFirst, It inLast) {
    if (preFirst == preLast && inFirst == inLast) return;

    // pre and in
    It inMid = find(inFirst, inLast, *preFirst);
    It preMid = advance(preFirst, distance(inFirst, inMid) + 1);

    // pre and post
    //It postMid = find(postFirst, --postLast, ++preFirst);
    //It preMid = find(preFirst, preLast, *postLast);

    // in and post
    //It inMid = find(inFirst, inLast, *--postLast);
    //It postMid = advance(postFirst, distance(inFirst, inMid) + 1);

    typedef Node<typename iterator_traits<It>::value_type> NodeType;
    NodeType* root = new NodeType;
    root->v = *preFirst;
    root->left = buildTree(++preFirst, preMid, inFirst, inMid);
    root->right = buildTree(preMid, preLast, ++inMid, inLast);
    return root;
}

template<class It>
Node<typename iterator_traits<It>::value_type>*
build(It preFirst, It preLast, It inFirst, It inLast) {
    if (!buildable(preFirst, preLast, inFirst, inLast)) return 0;
    return buildTree(preFirst, preLast, inFirst, inLast);
}