Time: 117 ms (48.02%), Space: 14.6 MB (88.89%) - LeetHub

2026-06-13 03:27:07 +00:00 · 2023-01-18 22:30:31 -05:00
parent cc463a3ea8
commit 870ed27f67
1 changed files with 51 additions and 0 deletions
@@ -0,0 +1,51 @@
+# 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]:
+        # Gets next node in traversal (and removes it from the string)
+        #   Returns (value, depth)
+        def getNextNode() -> tuple[int, int]:
+            nonlocal traversal
+            
+            # Find depth
+            depth = 0
+            while traversal[0] == "-":
+                depth += 1
+                traversal = traversal[1:]
+                
+            # Find value
+            value = 0
+            while len(traversal) > 0 and traversal[0] != "-":
+                value = (value * 10) + int(traversal[0])
+                traversal = traversal[1:]
+                
+            return (value, depth)
+        
+        # Base Case: empty tree
+        if len(traversal) == 0:
+            return None
+        
+        root = TreeNode(getNextNode()[0])
+        currentRoute = [root]
+        
+        while len(traversal) > 0:
+            nodePair = getNextNode()
+            
+            # Go deeper
+            if nodePair[1] == len(currentRoute):
+                currentRoute[-1].left = TreeNode(nodePair[0])
+                currentRoute.append(currentRoute[-1].left)
+                continue
+                
+            # Go back up the tree
+            while nodePair[1] < len(currentRoute):
+                currentRoute.pop()
+            
+            currentRoute[-1].right = TreeNode(nodePair[0])
+            currentRoute.append(currentRoute[-1].right)
+            
+        return root