---
tags: leetcode
---

# [285. Inorder Successor in BST](https://leetcode.com/problems/inorder-successor-in-bst/)
---
# My Solution
## Solution 1: Use BST property
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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 *successor = nullptr;
        while (root != nullptr) {
            if (root->val <= p->val) {
                root = root->right;
            } else {
                successor = root;
                root = root->left;
            }
        }

        return successor;
    }
};

### Time Complexity
O(H), H is the height of BST.
### Space Complexity
O(1)
## Solution 2: DFS (iteration with stack, inorder traversal)
### The Key Idea for Solving This Coding Question
### C++ Code

/**
 * 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) {
        stack<TreeNode *> st;
        bool isFound = false;

        while (root != nullptr || !st.empty()) {
            if (root != nullptr) {
                st.push(root);
                root = root->left;
            } else {
                root = st.top();
                st.pop();

                if (p == root) {
                    isFound = true;
                } else if (isFound == true) {
                    return root;
                }

                root = root->right;
            }
        }

        return nullptr;
    }
};

### Time Complexity
O(n), n is the total number of nodes in BST.
### Space Complexity
O(H), H is the height of BST.
# Miscellaneous
<!--
# Test Cases

[2,1,3]
1

[5,3,6,2,4,null,null,1]
6

[2,null,3]
2

-->