Created
September 23, 2018 20:55
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Inorder Successor in BST
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| """ | |
| Description: | |
| Given a binary search tree and a node in it, find the in-order successor of that node | |
| in the BST. If the given node has no in-order successor in the tree, return null. | |
| Examples: | |
| Input: root = [2,1,3], p = 1 | |
| 2 | |
| / \ | |
| 1 3 | |
| Output: 2 | |
| Input: root = [5,3,6,2,4,null,null,1], p = 6 | |
| 5 | |
| / \ | |
| 3 6 | |
| / \ | |
| 2 4 | |
| / | |
| 1 | |
| Output: null | |
| """ | |
| class TreeNode: | |
| def __init__(self, x): | |
| self.val = x | |
| self.left = None | |
| self.right = None | |
| def inorderSuccessor(root, p): | |
| """ | |
| :type root: TreeNode | |
| :type p: TreeNode | |
| :rtype: TreeNode | |
| """ | |
| if not root or not p: return None | |
| if root.val > p.val: | |
| return inorderSuccessor(root.left, p) or root | |
| return inorderSuccessor(root.right, p) |
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