Sync LeetCode submission Runtime - 62 ms (24.45%), Memory - 21 MB (41.74%)

Author: jimit105

0285-inorder-successor-in-bst/README.md

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+<p>Given the <code>root</code> of a binary search tree and a node <code>p</code> in it, return <em>the in-order successor of that node in the BST</em>. If the given node has no in-order successor in the tree, return <code>null</code>.</p>
+
+<p>The successor of a node <code>p</code> is the node with the smallest key greater than <code>p.val</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/01/23/285_example_1.PNG" style="width: 122px; height: 117px;" />
+<pre>
+<strong>Input:</strong> root = [2,1,3], p = 1
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> 1&#39;s in-order successor node is 2. Note that both p and the return value is of TreeNode type.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2019/01/23/285_example_2.PNG" style="width: 246px; height: 229px;" />
+<pre>
+<strong>Input:</strong> root = [5,3,6,2,4,null,null,1], p = 6
+<strong>Output:</strong> null
+<strong>Explanation:</strong> There is no in-order successor of the current node, so the answer is <code>null</code>.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
+	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
+	<li>All Nodes will have unique values.</li>
+</ul>

0285-inorder-successor-in-bst/solution.py

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+# Approach 2: Using BST properties
+
+# Time: O(n)
+# Space: O(1)
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> Optional[TreeNode]:
+        successor = None
+
+        while root:
+            if p.val >= root.val:
+                root = root.right
+            else:
+                successor = root
+                root = root.left
+
+        return successor