边工作边刷题：70天一遍leetcode: day 85-1

Inorder Successor in BST

要点：这题要注意的是如果不是BST，没法从树结构上从root向那边找p，只能遍历。而根据BST，可以只走正确方向

• 如果不检查right子树，可以从root到下，但invariant是root!=null。而检查右子树，invariant可以是root!=p

错误点：

• 不是找到某个>p.val，而是要找到最接近的p.val：所以loop终止条件是直到p==root or root is None，过程中只要>p.val就记录successor：这个过程的前提是p.right is None
• 别忘了loop

https://repl.it/CfBm/1

# Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

# Note: If the given node has no in-order successor in the tree, return null.

# Hide Company Tags Pocket Gems Microsoft Facebook
# Hide Tags Tree
# Hide Similar Problems (M) Binary Tree Inorder Traversal (M) Binary Search Tree Iterator

# 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
        """
        if p.right:
            p = p.right
            while p.left:
                p=p.left
            return p
        else:
            succ = None
            while root!=p:
                if p.val<root.val:
                    succ = root
                    root = root.left
                else:
                    root = root.right
            return succ