根据中序遍历和后序遍历树构造二叉树
注意事项
你可以假设树中不存在相同数值的节点
样例
给出树的中序遍历： [1,2,3] 和后序遍历： [1,3,2]
返回如下的树：

2

/ \

1 3
题目链接：http://www.lintcode.com/zh-cn/problem/construct-binary-tree-from-inorder-and-postorder-traversal/

后序遍历的最后一个数为二叉树的根结点，在中序遍历中找到这个“根结点”，将中序遍历分为两部分，左半部分为根的左子树，右半部分为根的右子树。再把这两部分分别当成一个完整的树，于是使用递归来构造这棵二叉树：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 

class Solution {
    /**
     *@param inorder : A list of integers that inorder traversal of a tree
     *@param postorder : A list of integers that postorder traversal of a tree
     *@return : Root of a tree
     */
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // write your code here
        if (inorder.size() == 0) return NULL;
        return build(inorder,postorder,0,inorder.size() - 1,0,postorder.size() - 1);
    }
    TreeNode *build(vector<int> &inorder,vector<int> postorder,int startin,int endin,int startpost,int endpost) {
        if (startin > endin) return NULL;
        TreeNode *root = new TreeNode(postorder[endpost]);
        int divide = 0;
        while (divide <= endin && inorder[divide] != root->val) divide++;
        int offset = divide - startin - 1;
        root->left = build(inorder,postorder,startin,startin + offset,startpost,startpost + offset);
        root->right = build(inorder,postorder,divide + 1,endin,startpost + offset + 1,endpost - 1);
        return root;
    }
};