Populating Next Right Pointers in Each Node II

Author: anantkaushik

117-populating-next-right-pointers-in-each-node-ii.py

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+"""
+Problem Link: https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
+
+Given a binary tree
+struct Node {
+  int val;
+  Node *left;
+  Node *right;
+  Node *next;
+}
+Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be 
+set to NULL.
+Initially, all next pointers are set to NULL.
+
+Follow up:
+You may only use constant extra space.
+Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.
+ 
+Example 1:
+Input: root = [1,2,3,4,5,null,7]
+Output: [1,#,2,3,#,4,5,7,#]
+Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its 
+next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, 
+with '#' signifying the end of each level.
+ 
+Constraints:
+The number of nodes in the given tree is less than 6000.
+-100 <= node.val <= 100
+"""
+"""
+# 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, parent=None):
+        if not root:
+            return
+        cur1 = cur2 = None
+        while parent:
+            
+            if not parent.left and not parent.right:
+                parent = parent.next
+                continue
+                
+            if not cur1:
+                cur1 = parent.left or parent.right
+                if cur1 == parent.right:
+                    parent = parent.next
+                    
+            elif not cur2:
+                if cur1 == parent.right or cur1 == parent.left and not parent.right:
+                    parent = parent.next
+                    continue
+                if cur1 == parent.left:
+                    cur2 = parent.right
+                else:
+                    cur2 = parent.left or parent.right
+                    
+            else:
+                cur1.next = cur2
+                cur1, cur2 = cur2, None
+            
+        self.dfs(root.left, root)
+        self.dfs(root.right, root)
+        
+        
+class Solution1:
+    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)