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Space - Chelsea #20
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Space - Chelsea #20
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Original file line number | Diff line number | Diff line change |
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@@ -14,42 +14,172 @@ class Tree | |
attr_reader :root | ||
def initialize | ||
@root = nil | ||
@inOrderArray = [] | ||
end | ||
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# Time Complexity: | ||
# Space Complexity: | ||
# Time Complexity: log n because it is binary to find where to add the node | ||
# Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
def add(key, value) | ||
raise NotImplementedError | ||
if @root == nil | ||
@root = TreeNode.new(key, value) | ||
return | ||
end | ||
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current = @root | ||
add_helper(current, key, value) | ||
end | ||
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# Time Complexity: | ||
# Space Complexity: | ||
def add_helper(current, key, value) | ||
if current.key >= key | ||
if current.left | ||
current = current.left | ||
add_helper(current, key, value) | ||
else | ||
newNode = TreeNode.new(key, value) | ||
current.left = newNode | ||
end | ||
elsif current.key < key | ||
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if current.right | ||
current = current.right | ||
add_helper(current, key, value) | ||
else | ||
newNode = TreeNode.new(key, value) | ||
current.right = newNode | ||
end | ||
end | ||
end | ||
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# Time Complexity: log n due to the binary search | ||
# Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
def find(key) | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 |
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raise NotImplementedError | ||
if !@root | ||
return nil | ||
end | ||
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current = @root | ||
return find_helper(current, key) | ||
end | ||
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# Time Complexity: | ||
# Space Complexity: | ||
def find_helper(current, searchKey) | ||
if current.key == searchKey | ||
return current.value | ||
end | ||
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if current.key > searchKey | ||
current = current.left | ||
return find_helper(current, searchKey) | ||
elsif current.key < searchKey | ||
current = current.right | ||
return find_helper(current, searchKey) | ||
end | ||
end | ||
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# Time Complexity: O(n) because it has to go through every node. | ||
# Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
def inorder | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 |
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raise NotImplementedError | ||
if @root == nil | ||
return [] | ||
end | ||
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current = @root | ||
getInOrder(current) | ||
return @inOrderArray | ||
end | ||
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# Time Complexity: | ||
# Space Complexity: | ||
def getInOrder(node) | ||
if node.left != nil | ||
getInOrder(node.left) | ||
end | ||
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@inOrderArray.push({:key=>node.key, :value=>node.value}) | ||
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if node.right != nil | ||
getInOrder(node.right) | ||
end | ||
end | ||
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# Time Complexity: O(n) because it has to go through every node. | ||
# Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
def preorder | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. The space complexity for these traversals is O(n) because we're building an array. With recursion it would be O(n + n) = O(2n) = O(n) or average case O(n + log n) = O(n) |
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raise NotImplementedError | ||
if @root == nil | ||
return [] | ||
end | ||
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current = @root | ||
getPreOrder(current) | ||
return @inOrderArray | ||
end | ||
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# Time Complexity: | ||
# Space Complexity: | ||
def getPreOrder(node) | ||
@inOrderArray.push({:key=>node.key, :value=>node.value}) | ||
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if node.left != nil | ||
getPreOrder(node.left) | ||
end | ||
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if node.right != nil | ||
getPreOrder(node.right) | ||
end | ||
end | ||
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# Time Complexity: O(n) because it has to go through every node. | ||
# Space Complexity: worst case O(n) because we hit every node, better average case like in a balanced tree it would be O(log n) | ||
def postorder | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 With the earlier note on the traversals. |
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raise NotImplementedError | ||
if @root == nil | ||
return [] | ||
end | ||
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current = @root | ||
getPostOrder(current) | ||
return @inOrderArray | ||
end | ||
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# Time Complexity: | ||
# Space Complexity: | ||
def getPostOrder(node) | ||
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if node.left != nil | ||
getPostOrder(node.left) | ||
end | ||
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if node.right != nil | ||
getPostOrder(node.right) | ||
end | ||
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@inOrderArray.push({:key=>node.key, :value=>node.value}) | ||
end | ||
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# Time Complexity: O(n) because we hit eery node | ||
# Space Complexity: worst case O(n) because we hit every node better average case like in a balanced tree it would be O(log n) | ||
def height | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 👍 |
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raise NotImplementedError | ||
if @root == nil | ||
return 0 | ||
end | ||
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current = @root | ||
height = 1 | ||
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height = getHeight(current, height) | ||
return height | ||
end | ||
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def getHeight(node, height) | ||
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left = height | ||
right = height | ||
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if node.left != nil | ||
left = getHeight(node.left, height +1) | ||
end | ||
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if node.right != nil | ||
right = getHeight(node.right, height + 1) | ||
end | ||
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if left >= right | ||
return left | ||
else | ||
return right | ||
end | ||
end | ||
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# Optional Method | ||
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👍