Sync LeetCode submission Runtime - 1 ms (93.72%), Memory - 21.4 MB (25.69%)

Author: jimit105

0663-equal-tree-partition/README.md

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+<p>Given the <code>root</code> of a binary tree, return <code>true</code><em> if you can partition the tree into two trees with equal sums of values after removing exactly one edge on the original tree</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/split1-tree.jpg" style="width: 500px; height: 204px;" />
+<pre>
+<strong>Input:</strong> root = [5,10,10,null,null,2,3]
+<strong>Output:</strong> true
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/split2-tree.jpg" style="width: 277px; height: 302px;" />
+<pre>
+<strong>Input:</strong> root = [1,2,10,null,null,2,20]
+<strong>Output:</strong> false
+<strong>Explanation:</strong> You cannot split the tree into two trees with equal sums after removing exactly one edge on the tree.
+</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>

0663-equal-tree-partition/solution.py

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+# Approach: Depth First Search
+
+# 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
+
+class Solution:
+    def checkEqualTree(self, root: Optional[TreeNode]) -> bool:
+        seen = []
+
+        def sum_(node):
+            if not node:
+                return 0
+            seen.append(sum_(node.left) + sum_(node.right) + node.val)
+            return seen[-1]
+
+        total = sum_(root)
+        seen.pop()
+        return total / 2.0 in seen
+