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chore(book/exercises): bst exercises after traversals
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‎book/D-interview-questions-solutions.asc

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@@ -83,9 +83,9 @@ A better solution is to eliminate the 2nd for loop and only do one pass.
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Algorithm:
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- Do one pass through all the prices
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- Keep track of the minimum price seen so far.
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- calculate `profit = currentPrice - minPriceSoFar`
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- Keep track of the maximun profit seen so far.
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-- Keep track of the minimum price seen so far.
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-- calculate `profit = currentPrice - minPriceSoFar`
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-- Keep track of the maximun profit seen so far.
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- Return maxProfit.
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[source, javascript]
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- Have a pointer for each list
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- While there's a pointer that is not null, visite them
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- Compare each list node's value and take the smaller one.
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- Advance the pointer of the taken node to the next one.
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-- Compare each list node's value and take the smaller one.
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-- Advance the pointer of the taken node to the next one.
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*Implementation*:
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@@ -163,8 +163,8 @@ A better way to solve this problem is iterating over each character on both list
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- Set a pointer to iterate over each node in the lists.
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- For each node, have an index (starting at zero) and compare if both lists have the same data.
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- When the index reaches the last character on the current node, we move to the next node.
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- If we found that a character from one list doesn't match the other, we return `false`.
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-- When the index reaches the last character on the current node, we move to the next node.
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-- If we found that a character from one list doesn't match the other, we return `false`.
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*Implementation*:
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- Create a mapping for each opening bracket to its closing counterpart.
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- Iterate through the string
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- When we found an opening bracket, insert the corresponding closing bracket into the stack.
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- When we found a closing bracket, pop from the stack and make sure it corresponds to the current character.
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-- When we found an opening bracket, insert the corresponding closing bracket into the stack.
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-- When we found a closing bracket, pop from the stack and make sure it corresponds to the current character.
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- Check the stack is empty. If there's a leftover, it means that something didn't close properly.
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*Implementation*:
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*Algorithm*:
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- Traverse the daily temperatures backward
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- Push each temperature to a stack.
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- While the current temperature is larger than the one at the top of the stack, pop it.
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- If the stack is empty, then there's no warmer weather ahead, so it's 0.
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- If the stack has an element, calculate the index delta.
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-- Push each temperature to a stack.
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-- While the current temperature is larger than the one at the top of the stack, pop it.
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-- If the stack is empty, then there's no warmer weather ahead, so it's 0.
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-- If the stack has an element, calculate the index delta.
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*Implementation*:
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@@ -311,8 +311,8 @@ This game is perfect to practice working with Queues. There are at least two opp
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- If the new location has food, remove that eaten food from its queue and place the next food on the map (if any).
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- If the new position doesn't have food, remove the tail of the snake since it moved.
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- If the snake new position hits itself, game over (return -1). To make this check, we have 2 options:
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- Queue: we can visit all the elements on the snake's queue (body) and check if a new position collides. That's `O(n)`
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- Set: we can maintain a `set` with all the snake locations, so the check is `O(1)`.
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-- Queue: we can visit all the elements on the snake's queue (body) and check if a new position collides. That's `O(n)`
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-- Set: we can maintain a `set` with all the snake locations, so the check is `O(1)`.
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- Move the snake's head to a new location (enqueue)
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- Return the score (snake's length - 1);
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@@ -337,10 +337,11 @@ As you can see, we opted for using a set to trade speed for memory.
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=== Solutions for Binary Tree Questions
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(((Interview Questions Solutions, Binary Tree)))
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:leveloffset: -1
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leveloffset: -1
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[#binary-tree-q-diameter-of-binary-tree]
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include::content/part03/tree-intro.asc[tag=binary-tree-q-diameter-of-binary-tree]
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include::content/part03/binary-search-tree-traversal.asc[tag=binary-tree-q-diameter-of-binary-tree]
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We are asked to find the longest path on a binary tree that might or might not pass through the root node.
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[#binary-tree-q-binary-tree-right-side-view]
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include::content/part03/tree-intro.asc[tag=binary-tree-q-binary-tree-right-side-view]
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include::content/part03/binary-search-tree-traversal.asc[tag=binary-tree-q-binary-tree-right-side-view]
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The first thing that might come to mind when you have to visit a tree, level by level, is BFS.
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We can visit the tree using a Queue and keep track when a level ends, and the new one starts.
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.There are several ways to solve this problem using BFS. Here are some ideas:
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- *1 Queue + Sentinel node*: we can use a special character in the `Queue` like `'*'` or `null` to indicate the level's change. So, we would start something like this `const queue = new Queue([root, '*']);`.
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- *2 Queues*: using a "special" character might be seen as hacky, so you can also opt to keep two queues: one for the current level and another for the next level.
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- *1 Queue + size tracking*: we track the Queue's `size` before the children are enqueued. That way, we know where the current level ends. We are going to implement this one.
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- *1 Queue + size tracking*: we track the Queue's `size` before the children are enqueued. That way, we know where the current level ends.
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We are going to implement BFS with "1 Queue + size tracking", since it's arguably the most elegant.
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*Algorithm*:
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- Enqueue root
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- While the queue has an element
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- Check the current size of the queue
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- Dequeue only `size` times, and for each dequeued node, enqueue their children.
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- Check if the node is the last one in its level and add it to the answer.
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-- Check the current size of the queue
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-- Dequeue only `size` times, and for each dequeued node, enqueue their children.
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-- Check if the node is the last one in its level and add it to the answer.
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*Implementation*:
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‎book/content/part03/binary-search-tree-traversal.asc

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----
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Post-order traverval will return `3, 4, 5, 15, 40, 30, 10`.
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==== Practice Questions
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(((Interview Questions, Binary Tree)))
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// tag::binary-tree-q-diameter-of-binary-tree[]
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===== Binary Tree Diameter
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*BT-1*) _Find the diameter of a binary tree. A tree's diameter is the longest possible path from two nodes (it doesn't need to include the root). The length of a diameter is calculated by counting the number of edges on the path._
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// end::binary-tree-q-diameter-of-binary-tree[]
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// _Seen in interviews at: Facebook, Amazon, Google_
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// Example 1:
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// [graphviz, tree-diameter-example-1, png]
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// ....
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// graph G {
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// 1 -- 2 [color=red]
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// 1 -- 3 [color=red]
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// 2 -- 4
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// 2 -- 5 [color=red]
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// }
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// ....
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// Example 2:
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// [graphviz, tree-diameter-example-2, png]
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// ....
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// graph G {
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// 1
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// 2
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// 3
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// 4
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// 5
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// 6
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// "null" [color=white, fontcolor = white]
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// "null." [color=white, fontcolor = white]
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// 7
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// 8
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// "null.." [color=white, fontcolor = white]
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// "null..." [color=white, fontcolor = white]
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// 9
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// 1 -- 2
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// 1 -- 3
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// 3 -- 4 [color=red]
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// 3 -- 5 [color=red]
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// 4 -- 6 [color=red]
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// 4 -- "null." [color=white]
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// 5 -- "null" [color=white]
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// 5 -- 7 [color=red]
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// 6 -- 8 [color=red]
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// 6 -- "null.." [color=white]
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// 7 -- "null..." [color=white]
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// 7 -- 9 [color=red]
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// }
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// ....
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Examples:
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[source, javascript]
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----
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/* 1
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/ \
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2 3
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/ \
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4 5 */
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diameterOfBinaryTree(toBinaryTree([1,2,3,4,5])); // 3
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// For len 3, the path could be 3-1-2-5 or 3-1-2-4
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/* 1
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/ \
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2 3
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/ \
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4 5
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/ \
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6 7
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/ \
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8 9 */
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const array = [1,2,3,null,null,4,5,6,null,null,7,8,null,null,9];
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const tree = BinaryTreeNode.from(array);
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diameterOfBinaryTree(tree); // 6 (path: 8-6-4-3-5-7-9)
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----
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Starter code:
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[source, javascript]
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----
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include::../../interview-questions/diameter-of-binary-tree.js[tags=description;placeholder]
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----
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_Solution: <<binary-tree-q-diameter-of-binary-tree>>_
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// tag::binary-tree-q-binary-tree-right-side-view[]
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===== Binary Tree from right side view
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*BT-2*) _Imagine that you are viewing the tree from the right side. What nodes would you see?_
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// end::binary-tree-q-binary-tree-right-side-view[]
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// _Seen in interviews at: Facebook, Amazon, ByteDance (TikTok)._
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Examples:
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[source, javascript]
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----
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/*
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1 <- 1 (only node on level)
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/ \
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2 3 <- 3 (3 is the rightmost)
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\
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4 <- 4 (only node on level) */
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rightSideView(BinaryTreeNode.from([1, 2, 3, null, 4])); // [1, 3, 4]
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rightSideView(BinaryTreeNode.from([])); // []
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rightSideView(BinaryTreeNode.from([1, 2, 3, null, 5, null, 4, 6])); // [1, 3, 4, 6]
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----
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Starter code:
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[source, javascript]
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----
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include::../../interview-questions/binary-tree-right-side-view.js[tags=description;placeholder]
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----
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_Solution: <<binary-tree-q-binary-tree-right-side-view>>_
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‎book/content/part03/tree-intro.asc

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Heap is better at finding max or min values in constant time *O(1)*, while a balanced BST is good a finding any element in *O(log n)*. Heaps are often used to implement priority queues while BST is used when you need every value sorted.
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****
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indexterm:[Runtime, Logarithmic]
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==== Practice Questions
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(((Interview Questions, Binary Tree)))
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// tag::binary-tree-q-diameter-of-binary-tree[]
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===== Binary Tree Diameter
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*BT-1*) _Find the diameter of a binary tree. A tree's diameter is the longest possible path from two nodes (it doesn't need to include the root). The length of a diameter is calculated by counting the number of edges on the path._
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// end::binary-tree-q-diameter-of-binary-tree[]
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// _Seen in interviews at: Facebook, Amazon, Google_
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// Example 1:
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// [graphviz, tree-diameter-example-1, png]
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// ....
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// graph G {
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// 1 -- 2 [color=red]
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// 1 -- 3 [color=red]
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// 2 -- 4
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// 2 -- 5 [color=red]
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// }
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// ....
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// Example 2:
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// [graphviz, tree-diameter-example-2, png]
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// ....
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// graph G {
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// 1
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// 2
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// 3
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// 4
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// 5
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// 6
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// "null" [color=white, fontcolor = white]
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// "null." [color=white, fontcolor = white]
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// 7
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// 8
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// "null.." [color=white, fontcolor = white]
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// "null..." [color=white, fontcolor = white]
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// 9
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// 1 -- 2
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// 1 -- 3
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// 3 -- 4 [color=red]
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// 3 -- 5 [color=red]
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// 4 -- 6 [color=red]
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// 4 -- "null." [color=white]
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// 5 -- "null" [color=white]
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// 5 -- 7 [color=red]
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// 6 -- 8 [color=red]
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// 6 -- "null.." [color=white]
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// 7 -- "null..." [color=white]
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// 7 -- 9 [color=red]
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// }
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// ....
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Examples:
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[source, javascript]
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----
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/* 1
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2 3
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4 5 */
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diameterOfBinaryTree(toBinaryTree([1,2,3,4,5])); // 3
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// For len 3, the path could be 3-1-2-5 or 3-1-2-4
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/* 1
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2 3
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4 5
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6 7
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8 9 */
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const array = [1,2,3,null,null,4,5,6,null,null,7,8,null,null,9];
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const tree = BinaryTreeNode.from(array);
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diameterOfBinaryTree(tree); // 6 (path: 8-6-4-3-5-7-9)
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----
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Starter code:
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[source, javascript]
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----
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include::../../interview-questions/diameter-of-binary-tree.js[tags=description;placeholder]
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----
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_Solution: <<binary-tree-q-diameter-of-binary-tree>>_
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// tag::binary-tree-q-binary-tree-right-side-view[]
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===== Binary Tree from right side view
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*BT-2*) _Imagine that you are viewing the tree from the right side. What nodes would you see?_
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// end::binary-tree-q-binary-tree-right-side-view[]
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// _Seen in interviews at: Facebook, Amazon, ByteDance (TikTok)._
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Examples:
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[source, javascript]
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----
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/*
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1 <- 1 (only node on level)
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/ \
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2 3 <- 3 (3 is the rightmost)
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\
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4 <- 4 (only node on level) */
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rightSideView(BinaryTreeNode.from([1, 2, 3, null, 4])); // [1, 3, 4]
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rightSideView(BinaryTreeNode.from([])); // []
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rightSideView(BinaryTreeNode.from([1, 2, 3, null, 5, null, 4, 6])); // [1, 3, 4, 6]
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----
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Starter code:
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[source, javascript]
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----
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include::../../interview-questions/binary-tree-right-side-view.js[tags=description;placeholder]
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----
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_Solution: <<binary-tree-q-binary-tree-right-side-view>>_

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