[leetcode] 285. Inorder Successor in BST @ python_python inorder successor in bst-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/88647472

博客围绕LeetCode上二叉搜索树中序后继问题展开。介绍了两种解法，解法1是通过中序遍历找出p.val后面的节点，若没有则返回None；解法2采用递归方式。还提及每种解法对应的代码。

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原题

https://leetcode.com/problems/inorder-successor-in-bst/

解法1

中序遍历, 找出p.val后面的节点, 如果没有就返回None.

代码

# 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 inorderSuccessor(self, root, p):
        """
        :type root: TreeNode
        :type p: TreeNode
        :rtype: TreeNode
        """
        l = []
        def inOrder(root):
            if not root:
                return
            inOrder(root.left)
            l.append(root.val)
            inOrder(root.right)
            
        inOrder(root)
        i = l.index(p.val)
        return TreeNode(l[i+1]) if i< len(l)-1 else None

解法2

递归

代码

# 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 inorderSuccessor(self, root, p):
        """
        :type root: TreeNode
        :type p: TreeNode
        :rtype: TreeNode
        """
        # base case
        if not root:
            return None
        if p.val < root.val:
            return self.inorderSuccessor(root.left, p) or root
        return self.inorderSuccessor(root.right, p)