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623. Add One Row to Tree

Difficulty Topics

Description

Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.

Note that the root node is at depth 1.

The adding rule is:

  • Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur's left subtree root and right subtree root.
  • cur's original left subtree should be the left subtree of the new left subtree root.
  • cur's original right subtree should be the right subtree of the new right subtree root.
  • If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root's left subtree.

 

Example 1:

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

Example 2:

Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • The depth of the tree is in the range [1, 104].
  • -100 <= Node.val <= 100
  • -105 <= val <= 105
  • 1 <= depth <= the depth of tree + 1

Solution

add-one-row-to-tree.py
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def addOneRow(self, root: Optional[TreeNode], val: int, depth: int) -> Optional[TreeNode]:
        if depth == 1:
            newRoot = TreeNode(val)
            newRoot.left = root

            return newRoot

        d = 1
        dq = deque([root])

        while dq:
            N = len(dq)

            for _ in range(N):
                node = dq.popleft()

                if d + 1 == depth:
                    oLeft = node.left
                    oRight = node.right
                    newLeftNode = TreeNode(val)
                    newLeftNode.left = oLeft

                    newRightNode = TreeNode(val)
                    newRightNode.right = oRight

                    node.left = newLeftNode
                    node.right = newRightNode

                else:
                    for child in filter(None, (node.left, node.right)):
                        dq.append(child)

            if d + 1 == depth:
                break

            d += 1

        return root