Populating Next Right Pointers in Each Node

Author: anantkaushik

116-populating-next-right-pointers-in-each-node.py

@@ -33,15 +33,90 @@
  / \  / \
 4->5->6->7 -> NULL
 """
-# Definition for binary tree with next pointer.
-# class TreeLinkNode:
-#     def __init__(self, x):
-#         self.val = x
-#         self.left = None
-#         self.right = None
-#         self.next = None
-
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
 class Solution:
+    def connect(self, root: 'Node') -> 'Node':
+        if not root:
+            return
+        
+        self.dfs(root)
+        return root
+        
+    
+    def dfs(self, root, next_node=None):
+        if not root:
+            return
+        
+        root.next = next_node
+        self.dfs(root.left, root.right)
+        self.dfs(root.right, root.next.left if root.next else None)
+
+
+class Solution1:
+    def connect(self, root: 'Node') -> 'Node':
+        if not root:
+            return
+        
+        self.dfs(root)
+        return root
+        
+    
+    def dfs(self, root, parent=None):
+        if not root:
+            return
+        
+        cur = None
+        while parent:
+            parent.left.next = parent.right
+            if not cur:
+                cur = parent.right
+            else:
+                cur.next = parent.left
+                cur = parent.right
+            parent = parent.next
+            
+        self.dfs(root.left, root)
+
+class Solution2:
+    def connect(self, root: 'Node') -> 'Node':
+        if not root:
+            return
+        
+        self.dfs(root)
+        return root
+        
+    
+    def dfs(self, root, next_node=None):
+        if not root:
+            return
+        
+        root.next = next_node
+        
+        n1 = root.right
+        n2 = root.next.left if root.next else None
+        n3 = root.next.right if root.next else None 
+        next_node = n1 or n2 or n3
+        if root.left:
+            self.dfs(root.left, next_node)
+        if root.right:
+            if next_node == n1:
+                next_node = n2 or n3
+            elif next_node == n2:
+                next_node = n3
+            else:
+                next_node = None
+            self.dfs(root.right, next_node)
+        
+
+class Solution3:
     # @param root, a tree link node
     # @return nothing
     def connect(self, root):