285. Inorder Successor in BST

• Difficulty: Medium

• Topics: Tree

• Similar Questions:

  • Binary Tree Inorder Traversal

  • Binary Search Tree Iterator

  • Inorder Successor in BST II

Problem:

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

The successor of a node p is the node with the smallest key greater than p.val.

Example 1:

Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of TreeNode type.

Example 2:

Input: root = [5,3,6,2,4,null,null,1], p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is null.

Note:

1. If the given node has no in-order successor in the tree, return null.
2. It's guaranteed that the values of the tree are unique.

Solutions:

/**
 * 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) {
        if (root == NULL || p == NULL)  return NULL;
        if (root->val <= p->val) {
            return inorderSuccessor(root->right, p);
        } else {
            TreeNode* ret = inorderSuccessor(root->left, p);
            return ret == NULL ? root : ret;
        }
    }
};