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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转载自www.cnblogs.com/rosyYY/p/10968089.html