（树）根据中序后序构建二叉树

• 题目：根据中序和后序遍历构建二叉树
• 思路：利用递归加上分治的思想。先找到根节点的值，然后在根据中序遍历找到根节点的左右两边的值，然后在递归的处理左右两边的左右子树。这里的关键在于怎么处理递归的左右子树的范围，代码里面详细解释
• 代码：

  class Solution {
  public:
      TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
          vector<int>::size_type lenIn = inorder.size();
          vector<int>::size_type lenPost = postorder.size();
          return buildTree_Aux(inorder,0,lenIn-1,postorder,0,lenPost-1);
      }
       
      TreeNode *buildTree_Aux(vector<int> &inorder,int inB,int inE,
                              vector<int> &postorder,int poB,int poE) {
          if(inB > inE || poB > poE)
              return NULL;
          //在后序遍历中确定根节点
          TreeNode* root = new TreeNode(postorder[poE]);
          //在中序遍历中确定左右子树
          vector<int>::iterator iter = find(inorder.begin()+inB,inorder.begin()+inE,postorder[poE]);
          int index = iter - inorder.begin();//根节点的位置
  
          root->left = buildTree_Aux(inorder,inB,index-1,postorder,poB,poB+index-1-inB);
  　　    //这里postorder,poB,poB+index-1-inB这部分表示后序的左子树。pob是开始位置，index-1-inB是后序左子树的节点数的个数。poB+index-1-inB是后序的左子树的尾部
          root->right = buildTree_Aux(inorder,index+1,inE,postorder,poB+index-inB,poE-1);
  　　    //这里postorder,poB+index-inB,poE-1这部分表示后序的右子树。poB+index-inB右子树开始位置，poE-1右子树结束位置，去掉了根节点的值（最尾部）
          return root;
      }
  };

• 题目：根据前序和中序遍历构建二叉树
• 思路：类似
• 代码：

  /**
   * 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> &preorder, vector<int> &inorder) {
          vector<int>::size_type lenIn = inorder.size();
          vector<int>::size_type lenPre = preorder.size();
          return buildTree_Aux(preorder, 0, lenPre-1, inorder, 0, lenIn-1);
      }
       TreeNode *buildTree_Aux(vector<int> &preorder,int prb,int pre,
                               vector<int> &inorder,int inb,int ine) {
           if (prb > pre || inb > ine)
               return NULL;
           TreeNode *root = new TreeNode(preorder[prb]);
           vector<int>::iterator iter = find(inorder.begin()+inb,inorder.begin()+ine,preorder[pre]);
           int mid = iter - inorder.begin();
           root->left = buildTree_Aux(preorder, prb+1, prb+mid-inb, inorder, inb, mid-1);
           root->right = buildTree_Aux(preorder, prb+1+mid-inb, pre, inorder, mid+1, ine);
           return root;
       }
  };