Sync LeetCode submission Runtime - 22 ms (46.63%), Memory - 18.2 MB (96.19%)

Author: jimit105

1093-recover-a-tree-from-preorder-traversal/README.md

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+<p>We run a&nbsp;preorder&nbsp;depth-first search (DFS) on the <code>root</code> of a binary tree.</p>
+
+<p>At each node in this traversal, we output <code>D</code> dashes (where <code>D</code> is the depth of this node), then we output the value of this node.&nbsp; If the depth of a node is <code>D</code>, the depth of its immediate child is <code>D + 1</code>.&nbsp; The depth of the <code>root</code> node is <code>0</code>.</p>
+
+<p>If a node has only one child, that child is guaranteed to be <strong>the left child</strong>.</p>
+
+<p>Given the output <code>traversal</code> of this traversal, recover the tree and return <em>its</em> <code>root</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex1.png" style="width: 423px; height: 200px;" />
+<pre>
+<strong>Input:</strong> traversal = &quot;1-2--3--4-5--6--7&quot;
+<strong>Output:</strong> [1,2,5,3,4,6,7]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex2.png" style="width: 432px; height: 250px;" />
+<pre>
+<strong>Input:</strong> traversal = &quot;1-2--3---4-5--6---7&quot;
+<strong>Output:</strong> [1,2,5,3,null,6,null,4,null,7]
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex3.png" style="width: 305px; height: 250px;" />
+<pre>
+<strong>Input:</strong> traversal = &quot;1-401--349---90--88&quot;
+<strong>Output:</strong> [1,401,null,349,88,90]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the original tree is in the range <code>[1, 1000]</code>.</li>
+	<li><code>1 &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
+</ul>

1093-recover-a-tree-from-preorder-traversal/solution.py

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+# Approach 2: Iterative Approach with Stack (Single Pass)
+
+# 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 recoverFromPreorder(self, traversal: str) -> Optional[TreeNode]:
+        stack = []
+        index = 0
+
+        while index < len(traversal):
+            # Count no. of dashes
+            depth = 0
+            while index < len(traversal) and traversal[index] == '-':
+                depth += 1
+                index += 1
+
+            # Node value
+            value = 0
+            while index < len(traversal) and traversal[index].isdigit():
+                value = value * 10 + int(traversal[index])
+                index += 1
+
+            node = TreeNode(value)
+
+            while len(stack) > depth:
+                stack.pop()
+
+            if stack:
+                if stack[-1].left is None:
+                    stack[-1].left = node
+                else:
+                    stack[-1].right = node
+
+            stack.append(node)
+
+        return stack[0]
+