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.

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.

TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) { // time: O(n); space: O(n)
    if (!root) return root;
    if (root->val <= p->val) {
        return inorderSuccessor(root->right, p);
    } else {
        TreeNode* left = inorderSuccessor(root->left, p);
        return !left ? root : left;
    }
}

void helper(TreeNode* root, TreeNode* p, TreeNode*& prev, TreeNode*& succ) { // time: O(n); space: O(n)
    if (!root) return;
    helper(root->left, p, prev, succ);
    if (prev == p) succ = root;
    prev = root;
    helper(root->right, p, prev, succ);
}
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
    if (!root || !p) return root;
    TreeNode* prev = nullptr, *succ = nullptr;
    helper(root, p, prev, succ);
    return succ;
}

TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) { // time: O(n); space: O(n)
    stack<TreeNode*> st;
    TreeNode* cur = root;
    bool found = false;
    while (cur || !st.empty()) {
        while (cur) {
            st.push(cur);
            cur = cur->left;
        }
        cur = st.top(); st.pop();
        if (found) return cur;
        if (cur == p) found = true;
        cur = cur->right;
    }
    return nullptr;
}

TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) { // time: O(n); space: O(1)
    if (!root) return root;
    TreeNode* cur = root, *successor = nullptr;
    while (cur) {
        if (cur->val <= p->val) {
            cur = cur->right;
        } else {
            successor = cur;
            cur = cur->left;
        }
    }
    return successor;
}