LeetCode 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


已知二叉树的前序遍历这中序遍历，求二叉树，这个其实挺简单的，在前序遍历中取第一个元素，然后在中序遍历找到相应的元素，左边的为左子树，右边的为右子树，根据长度在前序遍历中找到相应左子树和右子树的前序遍历。然后就可以递归了。

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        
        vector<int> leftpre;
        vector<int> leftino;
        vector<int> rightpre;
        vector<int> rightino;
        
        if(preorder.empty()||inorder.empty())
            return NULL;
        int root = preorder[0]; int left=0;int right=0;int tag=0;
        for(int i=0;i<inorder.size();i++)
        {
            if(inorder[i]==root)
            {
                tag=1;
            }
            else if(tag==0)
            {left++;leftino.push_back(inorder[i]);}
            else
            {right++;rightino.push_back(inorder[i]);}
        }
        for(int i=1;i<left+1;i++)
        {
            leftpre.push_back(preorder[i]);
        }
        for(int i=left+1;i<preorder.size();i++)
        {
            rightpre.push_back(preorder[i]);
        }
        
        TreeNode* tree = new TreeNode(root);
       
        tree->left = buildTree(leftpre,leftino);
        tree->right = buildTree(rightpre,rightino);
        
        return tree;
        
    }
};