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Author: jimit105

0925-construct-binary-tree-from-preorder-and-postorder-traversal/README.md

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+<p>Given two integer arrays, <code>preorder</code> and <code>postorder</code> where <code>preorder</code> is the preorder traversal of a binary tree of <strong>distinct</strong> values and <code>postorder</code> is the postorder traversal of the same tree, reconstruct and return <em>the binary tree</em>.</p>
+
+<p>If there exist multiple answers, you can <strong>return any</strong> of them.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/07/24/lc-prepost.jpg" style="width: 304px; height: 265px;" />
+<pre>
+<strong>Input:</strong> preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
+<strong>Output:</strong> [1,2,3,4,5,6,7]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> preorder = [1], postorder = [1]
+<strong>Output:</strong> [1]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= preorder.length &lt;= 30</code></li>
+	<li><code>1 &lt;= preorder[i] &lt;= preorder.length</code></li>
+	<li>All the values of <code>preorder</code> are <strong>unique</strong>.</li>
+	<li><code>postorder.length == preorder.length</code></li>
+	<li><code>1 &lt;= postorder[i] &lt;= postorder.length</code></li>
+	<li>All the values of <code>postorder</code> are <strong>unique</strong>.</li>
+	<li>It is guaranteed that <code>preorder</code> and <code>postorder</code> are the preorder traversal and postorder traversal of the same binary tree.</li>
+</ul>

0925-construct-binary-tree-from-preorder-and-postorder-traversal/solution.py

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+# Approach: Recursion
+
+# Time: O(n^2)
+# 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 constructFromPrePost(self, preorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
+        if not preorder or not postorder:
+            return None
+
+        # Create root node from first element of preorder
+        root = TreeNode(preorder[0])
+
+        if len(preorder) == 1:
+            return root
+
+        # Find the left subtree root in postorder
+        # The second element in preorder is the root of left subtree
+        left_root_val = preorder[1]
+        left_root_index = postorder.index(left_root_val)
+
+        left_size = left_root_index + 1
+
+        left_preorder = preorder[1 : left_size + 1]
+        right_preorder = preorder[left_size + 1 :]
+
+        left_postorder = postorder[: left_size]
+        right_postorder = postorder[left_size : -1]
+
+        root.left = self.constructFromPrePost(left_preorder, left_postorder)
+        root.right = self.constructFromPrePost(right_preorder, right_postorder)
+
+        return root
+