leetcode-105-从前序与中序遍历构造二叉树

题目描述：

方法一：

# 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:
        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()

另：

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
        def helper(i,j):
            while i<j:
                root = TreeNode(preorder[0])
                del preorder[0]
                root.left = helper(i,idx_map[root.val])
                root.right = helper(idx_map[root.val]+1,j)
                return root
        idx_map = {val:idx for idx, val in enumerate(inorder)}
        return helper(0,len(inorder))

原文地址：https://www.cnblogs.com/oldby/p/11184886.html