leetcode题解：Construct Binary Tree from Inorder and Postorder Traversal（根据中序和后序遍历构造二叉树）

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
13         return creatTree(inorder.begin(),inorder.end(),postorder.begin(),postorder.end());
14     }
15 private:
16     template<typename InputIterator>
17     TreeNode *creatTree(InputIterator in_beg,InputIterator in_end,InputIterator post_beg,InputIterator post_end)
18     {
19         if(in_beg==in_end||post_beg==post_end) return nullptr; //空树
20         TreeNode *root=new TreeNode(*(post_end-1));
21         auto inRootPos=find(in_beg,in_end,root->val);//中序遍历中找到根节点,返回迭代指针
22         int leftlen=distance(in_beg,inRootPos);//中序遍历起点指针与找到的根节点指针的距离
23         root->left=creatTree(in_beg,inRootPos,post_beg,next(post_beg,leftlen));//递归构建左子数
24         root->right=creatTree(next(inRootPos),in_end,next(post_beg,leftlen),post_end-1);//递归构建右子树
25         return root;
26     }
27 };