116. Populating Next Right Pointers in Each Node
- Total Submissions: 287603
- Difficulty: Medium
- Contributors: Admin
Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *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.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
Code
# 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
class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
if root is None:
return
pre = root
while pre.left:
pre_bakup = pre
while pre:
pre.left.next = pre.right
if pre.next:
pre.right.next = pre.next.left
pre = pre.next
pre = pre_bakup.left
Idea
You traverse nodes in a zigzag-like path. For every node being visited, you set its left child’s next to right child. If the node itself has next node, then its right child should also point to the node’s next node’s left child.
Below is the traversal path of pre:
The idea uses O(1) space and O(N) time, where N is the number of nodes in the binary tree.
Reference: https://discuss.leetcode.com/topic/2202/a-simple-accepted-solution
Code (Using BFS, space complexity is O(N))
# 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
from collections import *
class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
if root is None:
return
q = deque([(root, 0)])
while q:
if len(q) > 1:
# if two nodes have the same level
if q[0][1] == q[1][1]:
q[0][0].next = q[1][0]
n, lvl = q.popleft()
if n.left and n.right:
q.append((n.left, lvl+1))
q.append((n.right, lvl+1))