[leetcode]Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

首先，后序周游序列的最后一个元素是根节点。那么需要在中序周游序列中找到这个元素的下标，这样就可以将两个序列分别分割开，用同样的方法找根节点的左右子节点。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return makeTree(0,inorder.size()-1,0,inorder.size()-1,&inorder,&postorder);
    }
    TreeNode *makeTree(int l1,int r1,int l2,int r2,vector<int> *inorder , vector<int> *postorder)
    {
        if(l1 > r1)
            return NULL; 
        TreeNode * root;
        if(l1 == r1)
        {
            root = new TreeNode(postorder->at(r2));
            return root;
        }
        root = new TreeNode(postorder->at(r2));
        int pos = l1;
        while(inorder->at(pos) != postorder->at(r2))
            pos++;
        if(pos - 1 - l1 >= 0)
            root->left = makeTree(l1,pos - 1,l2,l2 + pos -1 - l1,inorder,postorder);
        if(r1 - pos - 1 >= 0)
            root->right = makeTree(pos + 1,r1,r2 - r1 + pos,r2 - 1,inorder,postorder);
        return root;
    }
};