106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.
根据中根遍历和后跟遍历，求二叉树。

class Solution {
public:
    TreeNode *createTree(vector<int>&in,int l1,int r1,vector<int>&post,int l2,int r2){
        if(l1>r1)return NULL;
        int len1,root=post[r2];
        for(int i=l1;i<=r1;i++){
            if(in[i]==root){
                len1=i-l1;
                break;
            }
        }
        TreeNode *q=new TreeNode(root);
        q->left=createTree(in,l1,l1+len1-1,post,l2,l2+len1-1);
        q->right=createTree(in,l1+len1+1,r1,post,l2+len1,r2-1);
        return q;
    }
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if (inorder.size() == 0) return NULL;
        TreeNode *head = createTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
        return head;
    }
};