Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

题目解析：

跟上题一样，利用遍历的特性，从后向前找。

class Solution{
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.size() != postorder.size())
            return NULL;
        return generateSub(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
    }
    TreeNode *generateSub(vector<int> &inorder,int begin1,int end1, vector<int> &postorder,int begin2,int end2){
        if(begin1>end1 || begin2>end2)
            return NULL;
        int i;
        for(i = end1;i >= begin1;i--){
            if(inorder[i] == postorder[end2])
                break;
        }

        TreeNode *p = new TreeNode(postorder[end2]);
        p->right = generateSub(inorder,i+1,end1,postorder,end2-end1+i,end2-1);
        p->left = generateSub(inorder,begin1,i-1,postorder,begin2,end2-end1+i-1);

        return p;
    }
};