Sync LeetCode submission Runtime - 55 ms (14.43%), Memory - 20.8 MB (5.13%)

Author: jimit105

0501-find-mode-in-binary-search-tree/README.md

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+<p>Given the <code>root</code> of a binary search tree (BST) with duplicates, return <em>all the <a href="https://en.wikipedia.org/wiki/Mode_(statistics)" target="_blank">mode(s)</a> (i.e., the most frequently occurred element) in it</em>.</p>
+
+<p>If the tree has more than one mode, return them in <strong>any order</strong>.</p>
+
+<p>Assume a BST is defined as follows:</p>
+
+<ul>
+	<li>The left subtree of a node contains only nodes with keys <strong>less than or equal to</strong> the node&#39;s key.</li>
+	<li>The right subtree of a node contains only nodes with keys <strong>greater than or equal to</strong> the node&#39;s key.</li>
+	<li>Both the left and right subtrees must also be binary search trees.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/11/mode-tree.jpg" style="width: 142px; height: 222px;" />
+<pre>
+<strong>Input:</strong> root = [1,null,2,2]
+<strong>Output:</strong> [2]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> root = [0]
+<strong>Output:</strong> [0]
+</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>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

0501-find-mode-in-binary-search-tree/solution.py

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+# Approach 1: Count Frequency with Hash Map (DFS)
+
+# Time: O(n)
+# Space: O(n)
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+from collections import defaultdict
+
+class Solution:
+    def findMode(self, root: Optional[TreeNode]) -> List[int]:
+        def dfs(node, counter):
+            if not node:
+                return
+
+            counter[node.val] += 1
+            dfs(node.left, counter)
+            dfs(node.right, counter)
+
+        counter = defaultdict(int)
+        dfs(root, counter)
+        max_freq = max(counter.values())
+
+        ans = []
+        for key in counter:
+            if counter[key] == max_freq:
+                ans.append(key)
+
+        return ans
+