LeetCode--105--从前序与中序遍历序列构造二叉树（python）

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

class Solution:
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        def helper(in_left = 0, in_right = len(inorder)):
            nonlocal pre_idx
            # if there is no elements to construct subtrees
            if in_left == in_right:
                return None
            
            # pick up pre_idx element as a root
            root_val = preorder[pre_idx]
            root = TreeNode(root_val)

            # root splits inorder list
            # into left and right subtrees
            index = idx_map[root_val]

            # recursion 
            pre_idx += 1
            # build left subtree
            root.left = helper(in_left, index)
            # build right subtree
            root.right = helper(index + 1, in_right)
            return root
        
        # start from first preorder element
        pre_idx = 0
        # build a hashmap value -> its index
        idx_map = {val:idx for idx, val in enumerate(inorder)} 
        return helper()

 2019-9-20

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        if len(preorder)==0:
            return None
        if len(preorder)==1:
            return TreeNode(preorder[0])
        root = TreeNode(preorder[0])
        inorderL = inorder[:inorder.index(preorder[0])]
        inorderR = inorder[inorder.index(preorder[0])+1:]
        root.left = self.buildTree(preorder[1:inorder.index(preorder[0])+1],inorderL)
        root.right = self.buildTree(preorder[inorder.index(preorder[0])+1:],inorderR)
        return root

2019-11-20 14:49:13