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Author: jimit105

0236-lowest-common-ancestor-of-a-binary-tree/README.md

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+<p>Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.</p>
+
+<p>According to the <a href="https://en.wikipedia.org/wiki/Lowest_common_ancestor" target="_blank">definition of LCA on Wikipedia</a>: &ldquo;The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <b>a node to be a descendant of itself</b>).&rdquo;</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
+<pre>
+<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The LCA of nodes 5 and 1 is 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
+<pre>
+<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> root = [1,2], p = 1, q = 2
+<strong>Output:</strong> 1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>
+	<li><code>-10<sup>9</sup> &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
+	<li>All <code>Node.val</code> are <strong>unique</strong>.</li>
+	<li><code>p != q</code></li>
+	<li><code>p</code> and <code>q</code> will exist in the tree.</li>
+</ul>

0236-lowest-common-ancestor-of-a-binary-tree/solution.py

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+# Approach 1: Recursive
+
+# Time: O(n)
+# Space: O(n)
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+
+    def __init__(self):
+        self.ans = None
+
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+
+        def recurse_tree(curr_node) -> bool:
+            # If reached end of branch, return False
+            if not curr_node:
+                return False
+
+            left = recurse_tree(curr_node.left)
+            right = recurse_tree(curr_node.right)
+
+            # If the current node is one of p or q
+            mid = curr_node == p or curr_node == q
+
+            # If any two of the three flags (left, right, mid) become True
+            if mid + left + right >= 2:
+                self.ans = curr_node
+
+            # Return true if either of the three bool values is True
+            return mid or left or right
+
+        recurse_tree(root)
+        return self.ans
+