二叉树--通过前序和中序构造二叉树

105. Construct Binary Tree from Preorder and Inorder Traversal

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

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

For example, given

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

Return the following binary tree:

3
   / \
  9  20
    /  \
   15   7

class Solution
{
private:
  TreeNode* buildTree(vector<int>::iterator PreBegin, vector<int>::iterator PreEnd,
    vector<int>::iterator InBegin, vector<int>::iterator InEnd)
  {
    if (PreBegin == PreEnd)
    {
      return NULL;
    }

    int HeadValue = *PreBegin;
    TreeNode *HeadNode = new TreeNode(HeadValue);

    vector<int>::iterator LeftEnd = find(InBegin, InEnd, HeadValue);
    if (LeftEnd != InEnd)
    {
      HeadNode->left = buildTree(PreBegin + 1, PreBegin + (LeftEnd - InBegin) + 1,
        InBegin, LeftEnd);
    }

    HeadNode->right = buildTree(PreBegin + (LeftEnd - InBegin) + 1, PreEnd,
      LeftEnd + 1, InEnd);

    return HeadNode;
  }
public:
  TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder)
  {
    if (preorder.empty())
    {
      return NULL;
    }

    return buildTree(preorder.begin(), preorder.end(), inorder.begin(),
      inorder.end());

  }
};