根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

这个题想通了就不难了，前序遍历的第一个一定是根节点，然后在中序遍历中找到这个节点，这个节点左右又是本节点左右的中序遍历，前序遍历中左子树的数目和中序一定一样，所以前序遍历从第二个点开始的到最后划分为前后两个集合（根据中序遍历得到的左右子树数目），这就是左右子树的前序遍历集合。以此递归即可。

C++源代码：

/**
 * Definition for a binary tree node.
 * 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) {
        return buildTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
    }
    TreeNode* buildTree(vector<int>& preorder, int pLeft, int pRight, vector<int>& inorder, int iLeft, int iRight){
        if (pLeft > pRight || iLeft > iRight) return NULL;
        int i = 0;
        for(i=iLeft;i<=iRight;i++){
            if(preorder[pLeft]==inorder[i]) break;
        }
        TreeNode* cur = new TreeNode(preorder[pLeft]);
        cur->left = buildTree(preorder, pLeft + 1, pLeft+i-iLeft, inorder, iLeft, i-1);
        cur->right = buildTree(preorder, pLeft+i-iLeft+1, pRight, inorder, i+1, iRight);
        return cur;
    }
};

python3源代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        return self.buildTree2(preorder, 0, len(preorder)-1, inorder, 0, len(inorder)-1)
    
    def buildTree2(self, preorder, pLeft, pRight, inorder, iLeft, iRight):
        if pLeft > pRight or iLeft > iRight:
            return None
        i = 0
        for i in range(iRight+1):
            if preorder[pLeft]==inorder[i]:
                break;
        cur = TreeNode(preorder[pLeft])
        cur.left = self.buildTree2(preorder, pLeft+1, pLeft+i-iLeft, inorder, iLeft, i-1)
        cur.right = self.buildTree2(preorder, pLeft+i-iLeft+1, pRight, inorder, i+1, iRight)
        return cur