leetcode/117-populating-next-right-pointers-in-each-node-ii.py at master · anantkaushik/leetcode · GitHub

Name: leetcode/117-populating-next-right-pointers-in-each-node-ii.py at master · anantkaushik/leetcode · GitHub
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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
"""

classSolution:
defconnect(self, root: 'Node') ->'Node':
ifnotroot:
return

self.dfs(root)
returnroot


defdfs(self, root, parent=None):
ifnotroot:
return
cur1=cur2=None
whileparent:

ifnotparent.leftandnotparent.right:
parent=parent.next
continue

ifnotcur1:
cur1=parent.leftorparent.right
ifcur1==parent.right:
parent=parent.next

elifnotcur2:
ifcur1==parent.rightorcur1==parent.leftandnotparent.right:
parent=parent.next
continue
ifcur1==parent.left:
cur2=parent.right
else:
cur2=parent.leftorparent.right

else:
cur1.next=cur2
cur1, cur2=cur2, None

self.dfs(root.left, root)
self.dfs(root.right, root)


classSolution1:
defconnect(self, root: 'Node') ->'Node':
ifnotroot:
return

self.dfs(root)
returnroot


defdfs(self, root, next_node=None):
ifnotroot:
return

root.next=next_node

n1=root.right
n2=root.next.leftifroot.nextelseNone
n3=root.next.rightifroot.nextelseNone
next_node=n1orn2orn3
ifroot.left:
self.dfs(root.left, next_node)
ifroot.right:
ifnext_node==n1:
next_node=n2orn3
elifnext_node==n2:
next_node=n3
else:
next_node=None
self.dfs(root.right, next_node)