[LeetCode] 285. Inorder Successor in BST

Problem

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

Solution


class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        //successor must be larger then the node itself, so:
        //if p is in root.left, root can be the successor, null cannot be
        //if p is in root.right, root can not be the successor, null can be
        if (root == null) return null;
        if (root.val &lt;= p.val) {
            return inorderSuccessor(root.right, p);
        } else {
            TreeNode leftRes = inorderSuccessor(root.left, p);
            if (leftRes == null) return root;
            return leftRes;
        }
    }
}

原文地址：https://segmentfault.com/a/1190000017066251