leetcode-mid-Linked list- 105. Construct Binary Tree from Preorder and Inorder Traversal
mycode 43.86%
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution(object):
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
if len(inorder)==0 or len(preorder)==0:
return
pos = inorder.index(preorder[0])
root = TreeNode(preorder[0])
root.left = self.buildTree(preorder[1:pos+1], inorder[:pos])
root.right = self.buildTree(preorder[pos+1:],inorder[pos+1:])
return root
参考:
可能map的方法会比index快
class Solution(object):
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
if not inorder or not preorder:
return None
map = {val:idx for idx, val in enumerate(inorder)}
pos = map[preorder[0]]
root = TreeNode(preorder[0])
root.left = self.buildTree(preorder[1:pos+1], inorder[:pos])
root.right = self.buildTree(preorder[pos+1:],inorder[pos+1:])
return root
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