Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.9 MB (6.39%)

Author: jimit105

0199-binary-tree-right-side-view/README.md

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+<p>Given the <code>root</code> of a binary tree, imagine yourself standing on the <strong>right side</strong> of it, return <em>the values of the nodes you can see ordered from top to bottom</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [1,2,3,null,5,null,4]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[1,3,4]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2024/11/24/tmpd5jn43fs-1.png" style="width: 400px; height: 207px;" /></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [1,2,3,4,null,null,null,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[1,3,4,5]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2024/11/24/tmpkpe40xeh-1.png" style="width: 400px; height: 214px;" /></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [1,null,3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[1,3]</span></p>
+</div>
+
+<p><strong class="example">Example 4:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = []</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[]</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[0, 100]</code>.</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+</ul>

0199-binary-tree-right-side-view/solution.py

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+# Approach 3: BFS: One Queue + Level Size Measurements
+
+# Time: O(n)
+# Space: O(d), d = tree diameter
+
+from collections import deque
+
+# 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
+
+class Solution:
+    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
+        if root is None:
+            return []
+
+        queue = deque([root])
+        rightside = []
+
+        while queue:
+            level_length = len(queue)
+
+            for i in range(level_length):
+                node = queue.popleft()
+
+                if i == level_length - 1:
+                    rightside.append(node.val)
+
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+
+        return rightside
+