[LeetCode] 623. Add One Row to Tree

Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.

The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root’s left subtree.

Example 1:

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Input: A binary tree as following: 4 / \ 2 6 / \ / 3 1 5 v = 1 d = 2 Output: 4 / \ 1 1 / \ 2 6 / \ / 3 1 5

Example 2:

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Input: A binary tree as following: 4 / 2 / \ 3 1 v = 1 d = 3 Output: 4 / 2 / \ 1 1 / \ 3 1

Note:

1. The given d is in range [1, maximum depth of the given tree + 1].
2. The given binary tree has at least one tree node.

Solution

This is a insertion only with different rules. Here we need to consider three cases:

1. If we arrive at the depth d we insert two nodes with value v as the left and right node of the current node.
2. If the current node has a left subtree and a right subtree, then we perform a recursive call on its left subtree and right subtree
3. If the current node only has a left subtree(or a right subtree) the we perform a recursive call on it left(right) subtree and leave the other side empty.

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data BTree = Node Int BTree BTree | EmptyTree deriving (Show, Eq, Read)

addRow :: BTree -> Int -> Int -> BTree
addRow (Node i left right) v d | (d - 1) == 1 =
                                   Node i (Node v left EmptyTree) (Node v EmptyTree right)
                               | left /= EmptyTree && right /= EmptyTree =
                                   Node i (addRow left v (d - 1)) (addRow right v (d-1))
                               | left /= EmptyTree =
                                   Node i (addRow left v (d - 1)) right
                               | right /= EmptyTree =
                                   Node i left (addRow right v (d - 1))

Complexity Analysis

The worst time complexity and space complexity are both \(O(n)\) if we want to insert the value as the deepest row in the tree.