算法描述：

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

Note:

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

For example, given

inorder = [9,3,15,20,7]postorder = [9,15,7,20,3]

Return the following binary tree:

3   / \  9  20    /  \   15   7

解题思路：理解原理，注意细节。

TreeNode* buildTree(vector
     
      & inorder, vector
      
       & postorder) {        return helper(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);    }        TreeNode* helper(vector
       
        & inorder, int is, int ie, vector
        
         & postorder, int ps, int pe){        if(is > ie || ps > pe) return nullptr;        TreeNode* node = new TreeNode(postorder[pe]);        int pos = 0;        for(int i=is; i <= ie; i++){            if(postorder[pe]==inorder[i]){                pos = i;                break;            }        }        node -> left = helper(inorder, is, pos-1, postorder, ps, ps - (is-pos+1));        node -> right = helper(inorder, pos+1, ie, postorder, ps - (is-pos+1)+1, pe-1);        return node;    }