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[7/27] 114. Flatten Binary Tree to Linked List

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Given the root of a binary tree, flatten the tree into a "linked list":

  • The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
  • The "linked list" should be in the same order as a pre-order traversal of the binary tree.

 

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

 

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -100 <= Node.val <= 100

 

Follow up: Can you flatten the tree in-place (with O(1) extra space)?
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Runtime: 57 ms, faster than 53.19% of Python3 online submissions for Flatten Binary Tree to Linked List.
Memory Usage: 15.3 MB, less than 47.67% of Python3 online submissions for Flatten Binary Tree to Linked List.
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root: return
        if not root.left and not root.right: return root
        
        
        if root.left:
            left_leaf = self.flatten(root.left)
            right_tmp = root.right
            root.right = root.left
            root.left = None
            left_leaf.right= right_tmp
        
        if root.right: 
            return self.flatten(root.right)
        
        return left_leaf
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