285. Inorder Successor in BST

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, returnnull.

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

Thoughts: Utilize the binary search property of BST: during the traversal: if a node is about to traverse to its left. Then record the node as successor before doing so.

Code: T: O(h) -> O(logn)

Python

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

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        TreeNode* suc = NULL;
        while(root ){
            if(root-> val > p ->val) {
                suc = root;
                root = root -> left;
            }
            else root = root -> right;
        }

        return suc;
    }
};

Java

public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
    TreeNode succ = null;
    while (root != null) {
        if (p.val < root.val) {
            succ = root;
            root = root.left;
        }
        else
            root = root.right;
    }
    return succ;
}