LeetCode #117 Populating Next Right Pointers in Each Node II

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Problem

Given a binary tree

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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.
  • Recursive approach is fine, implicit stack space does not count as extra space for this problem.

Example

Given the following binary tree,

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     1
   /  \
  2    3
 / \    \
4   5    7

After calling your function, the tree should look like:

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     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \    \
4-> 5 -> 7 -> NULL

Solution1

Method: Breadth-first Search

Time Complexity:

Space Complexity:

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# 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 not root:
            return

        dq = collections.deque([(root, 1)])
        
        while dq:
            node , level = dq.popleft()
            
            if dq:
                if dq[0][1] == level:
                    node.next = dq[0][0]
                    
            if node.left:
                dq.append((node.left, level + 1))
            if node.right:
                dq.append((node.right, level + 1))

Solution2

Method: Depth-first Search

Time Complexity:

Space Complexity:

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https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/37828/O(1)-space-O(n)-complexity-Iterative-Solution