給定一棵樹的前序遍歷?preorder?與中序遍歷??inorder。請構造二叉樹并返回其根節點。

示例 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

示例 2:

Input: preorder = [-1], inorder = [-1] Output: [-1]

提示:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder?和?inorder?均無重復元素
• inorder?均出現在?preorder
• preorder?保證為二叉樹的前序遍歷序列
• inorder?保證為二叉樹的中序遍歷序列

class Solution:def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:n = len(preorder)m = len(inorder)if m != n:return Nonemapper = {}for index,num in enumerate(inorder):mapper[num] = index#中序遍歷的索引return self.build(preorder,0,n-1,mapper,0,m-1)def build(self,preorder,preLeft,preRight,mapper,inLeft,inRight):if preLeft>preRight or inLeft>inRight:return Noneroot_val = preorder[preLeft]#前序遍歷的根root = TreeNode(root_val,None,None)index = mapper.get(root.val)root.left = self.build(preorder,preLeft+1,index-inLeft+preLeft,mapper,inLeft,index-1)root.right = self.build(preorder,index-inLeft+preLeft+1,preRight,mapper,index+1,inRight)return root

?

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