leetcode 106. 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.

后序遍历的最后一个是根节点，将中序遍历分开成左右子树。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* helper(vector<int> &inorder, int begin1, int end1, vector<int> &postorder, int begin2, int end2)    {        if (begin2 > end2)            return NULL;         int lenleft = find(inorder.begin(),inorder.end(),postorder[end2])-inorder.begin()-begin1;                TreeNode *p = new TreeNode(postorder[end2]);        p->left = helper(inorder,begin1,begin1+lenleft-1,postorder,begin2,begin2+lenleft-1);        p->right = helper(inorder,begin1+lenleft+1,end1,postorder,begin2+lenleft,end2-1);        return p;    }        TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)     {        return helper(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);    }};