Next Greater Number BST

Given a BST node, return the node which has value just greater than the given node.

Given the tree

               100
              /   \
            98    102
           /  \
         96    99
          \
           97

Given 97, you should return the node corresponding to 98 as thats the value just greater than 97 in the tree.
If there are no successor in the tree ( the value is the largest in the tree, return NULL).

Assume that the value is always present in the tree.

Method 1:

Inorder traversal

Solution 1:

Time: O(n)
Space: O(n)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode getSuccessor(TreeNode a, int b) {
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode curr = a;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            if (!stack.isEmpty()) {
                curr = stack.pop();
                if (curr.val > b) {
                    return curr;
                }
                curr = curr.right;
            }
        }
        return null;
    }
}

Method 2:

Binary Search

Solution2:

Time: O(n)
Space: O(1)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode getSuccessor(TreeNode a, int b) {
        TreeNode next = null;
        TreeNode curr = a;
        while (curr != null) {
            if (curr.val > b) {
                next = curr;
                curr = curr.left;
            } else {
                curr = curr.right;
            }
        }
        return next;
    }
}