The problem is to find the maximum amount of money you can rob from a binary tree house structure, where adjacent houses cannot be robbed simultaneously. The approach involves traversing the binary tree recursively while maintaining two options for each node:

  1. Option 0: Maximum amount of money that can be robbed if the current node is robbed.
  2. Option 1: Maximum amount of money that can be robbed if the current node is not robbed.

Approach

  1. Define a recursive function travel(TreeNode root) that returns an array of size 2 representing the two options mentioned above.
  2. In the base case, if the current node is null, return [0, 0].
  3. Recursively call the travel function for the left and right subtrees to get the options for the left and right child nodes.
  4. Calculate the options for the current node:
  • Option 0: The sum of the current node's value, the maximum amount robbed from the left subtree without robbing its root (leftNodeChoices[1]), and the maximum amount robbed from the right subtree without robbing its root (rightNodeChoices[1]).
  • Option 1: The maximum of the sum of the maximum amount robbed from the left subtree (either robbing or not robbing its root) and the maximum amount robbed from the right subtree (either robbing or not robbing its root).
  1. Return the options array.

Complexity

  • Time complexity:

The time complexity of the solution is , where N is the number of nodes in the binary tree. This is because each node is visited once during the recursive traversal.

  • Space complexity:

The space complexity is also  due to the recursion stack, where N is the height of the binary tree. Additionally, the space complexity includes the arrays created for each node to store the options, but since there are 2 options per node, it remains .

Comments

Post a Comment

Popular posts from this blog

30. Substring with Concatenation of All Words

Problem Statement You are given a string s and an array of strings words. All the strings of words are of the same length. A concatenated substring in s is a substring that contains all the strings of any permutation of words concatenated. For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd", and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is not the concatenation of any permutation of words. Return the starting indices of all the concatenated substrings in s. You can return the answer in any order. Example 1: Input: s = "barfoothefoobarman", words = ["foo","bar"] Output: [0,9] Explanation: Since words.length == 2 and words[i].length == 3, the concatenated substring has to be of length 6. The substring starting at 0 is "barfoo". It is the concaten...
Read more

Lowest common ancestor in a binary Tree

Approach Define a recursive function rec that takes three parameters: root (the current node), p (first node to find the lowest common ancestor for), and q (second node to find the lowest common ancestor for). Check if the current node root is null or if it is either p or q . If any of these conditions is true, return the current node root as the lowest common ancestor. Recursively call the rec function for the left subtree of the current node and assign the result to a variable l . Recursively call the rec function for the right subtree of the current node and assign the result to a variable r . Check if both l and r are not null. If so, it means that p and q are found on different subtrees of the current node, and the current node root is their lowest common ancestor. Return the current node root . If l is null, it means that both p and q are on the right subtree (or not present in the tree). Return r . If none of the above conditions are met, it means ...
Read more

New start day 1 : 30/09/2020 (codeforces 71A) Way to long strings

 Question:  source (codeforces: https://codeforces.com/problemset/problem/71/A) Way too long strings   Sometimes some words like " localization " or " internationalization " are so long that writing them many times in one text is quite tiresome. Let's consider a word  too long if its length is  strictly more  than  10  characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this: we write down the first and the last letter of a word and between them, we write the number of letters between the first and the last letters. That number is in the decimal system and doesn't contain any leading zeroes. Thus, " localization " will be spelled as " l10n ", and " internationalization » will be spelled as " i18n ". You are suggested to automatize the process of changing the words with abbreviations. At that, all too long words should be replaced by the abbreviation and the words th...
Read more