109.md

Convert Sorted List to Binary Search Tree

Description

link

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Solution

注意最重要的就是如何寻找到最中间的那个node，用一个fast和slow就可以快速定位

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Code

O(n)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        # recursively
        if not head:
            return 
        if not head.next:
            return TreeNode(head.val)
        # here we get the middle point,
        # even case, like '1234', slow points to '2',
        # '3' is root, '12' belongs to left, '4' is right
        # odd case, like '12345', slow points to '2', '12'
        # belongs to left, '3' is root, '45' belongs to right
        slow, fast = head, head.next.next
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        # tmp points to root
        tmp = slow.next
        # cut down the left child
        slow.next = None
        root = TreeNode(tmp.val)
        root.left = self.sortedListToBST(head)
        root.right = self.sortedListToBST(tmp.next)
        return root