Leetcode 之Construct Binary Tree（52）

根据先序和中序构造二叉树、根据中序和后序构造二叉树，基础题，采用递归的方式解决，两题的方法类似。需要注意的是迭代器的用法。

 //先序和中序
      TreeNode *buildTree(vector<int>& preorder, vector<int>& inorder)
      {
          return buildTree(begin(preorder), end(preorder), begin(inorder), begin(inorder));
      }
      template<typename InputIterator>
      TreeNode *buildTree(InputIterator pre_first, InputIterator pre_last,
          InputIterator in_first, InputIterator in_last)
      {
          if (pre_first == pre_last)return nullptr;
          if (in_first == in_last)return nullptr;
          //先序第一个为根结点
          auto root = new TreeNode(*pre_first);
          //查找根结点在中序的位置，返回的是迭代器
          auto intRootPos = find(in_first, in_last, *pre_first);
          //得到根结点的左半部分
          auto leftSize = distance(in_first, intRootPos);
          //递归构造，注意去掉根结点
          root->left = buildTree(next(pre_first), next(pre_first, leftSize),
              in_first, next(in_first, leftSize));
          root->right = buildTree(next(pre_first, leftSize), pre_last, next(intRootPos), in_last);

          return root;
      }

//中序和后序
      TreeNode *buildTree1(vector<int>& inorder, vector<int>& postorder)
      {
          return buildTree1(begin(inorder), end(inorder), begin(postorder), begin(postorder));
      }
      template<typename InputIterator>
      TreeNode *buildTree1(InputIterator in_first, InputIterator in_last,
          InputIterator post_first, InputIterator post_last)
      {
          if (in_first == in_last)return nullptr;
          if (post_first == post_last)return nullptr;
          //后序最后一个为根结点
          const auto val = *prev(post_last)
          auto root = new TreeNode(val);
          //查找根结点在中序的位置，返回的是迭代器
          auto intRootPos = find(in_first, in_last, val);
          //得到根结点的左半部分
          auto leftSize = distance(in_first, intRootPos);
          //递归构造，注意去掉根结点
          root->left = buildTree(in_first, intRootPos,
              post_first, next(post_first, leftSize));
          root->right = buildTree(next(intRootPos), in_last, 
              next(post_first,leftSize), prev(post_last));

          return root;
      }