标签：-o arc find pre height success root amp nod
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.
Example 1:
Input: root = [2,1,3], p = 1
2
/ 1 3
Output: 2
Example 2:
Input: root = [5,3,6,2,4,null,null,1], p = 6
5
/ 3 6
/ 2 4
/
1
Output: null
M1: recursive
time: O(n), space: O(height)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { if(root == null) { return null; } if(p.val >= root.val) { return inorderSuccessor(root.right, p); } else { TreeNode left = inorderSuccessor(root.left, p); if(left != null) { return left; } else { return root; } } } }
M2: iterative
分两种情况考虑：p有无右子节点
如果p有右子节点，返回右子树的最左子节点；如果没有，从root开始按inorder遍历找successor
time: O(n), space: O(1)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { if(root == null) { return null; } if(p.right != null) { TreeNode cur = p.right; while(cur.left != null) { cur = cur.left; } return cur; } else { TreeNode s = root, t = null; while(s.val != p.val) { if(p.val <= s.val) { t = s; s = s.left; } else { s = s.right; } } return t; } } }
285. Inorder Successor in BST - Medium
标签：-o arc find pre height success root amp nod
原文地址：https://www.cnblogs.com/fatttcat/p/10204178.html
