09 binary trees
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The document is a detailed analysis of binary search trees (BST), outlining their concepts, properties, and key operations such as tree-search, tree-minimum, and tree-delete. It also discusses the time complexity of operations on BSTs, emphasizing the importance of balanced trees for optimal performance. Furthermore, the document includes exercises for further understanding of the topics covered.
![Images/cinvestav-
Binary search trees: Concepts
Definition
A binary search tree (BST) is a data structure where each node posses
three fields left, right and p.
They represent its left child, right child and parent.
In addition, each node has the field key.
Property
Let x be a node in a binary search tree. If y is a node in the left subtree of
x, then key[y] ≤ key[x]. Similarly, if y is a node in the right subree of x,
then key[x] ≤ key[y].
4 / 15](https://image.slidesharecdn.com/09binarytrees-150321103911-conversion-gate01/85/09-binary-trees-4-320.jpg)
![Images/cinvestav-
Binary search trees: Concepts
Definition
A binary search tree (BST) is a data structure where each node posses
three fields left, right and p.
They represent its left child, right child and parent.
In addition, each node has the field key.
Property
Let x be a node in a binary search tree. If y is a node in the left subtree of
x, then key[y] ≤ key[x]. Similarly, if y is a node in the right subree of x,
then key[x] ≤ key[y].
4 / 15](https://image.slidesharecdn.com/09binarytrees-150321103911-conversion-gate01/85/09-binary-trees-5-320.jpg)
09 binary trees
- 1. Analysis of Algorithms Binary Trees Andres Mendez-Vazquez August 26, 2014 1 / 15
- 2. Images/cinvestav- Outline 1 Binary search trees Binary search trees: Concepts Binary search trees: Operations 2 Excercises 2 / 15
- 3. Images/cinvestav- Outline 1 Binary search trees Binary search trees: Concepts Binary search trees: Operations 2 Excercises 3 / 15
- 4. Images/cinvestav- Binary search trees: Concepts Definition A binary search tree (BST) is a data structure where each node posses three fields left, right and p. They represent its left child, right child and parent. In addition, each node has the field key. Property Let x be a node in a binary search tree. If y is a node in the left subtree of x, then key[y] ≤ key[x]. Similarly, if y is a node in the right subree of x, then key[x] ≤ key[y]. 4 / 15
- 5. Images/cinvestav- Binary search trees: Concepts Definition A binary search tree (BST) is a data structure where each node posses three fields left, right and p. They represent its left child, right child and parent. In addition, each node has the field key. Property Let x be a node in a binary search tree. If y is a node in the left subtree of x, then key[y] ≤ key[x]. Similarly, if y is a node in the right subree of x, then key[x] ≤ key[y]. 4 / 15
- 6. Images/cinvestav- Outline 1 Binary search trees Binary search trees: Concepts Binary search trees: Operations 2 Excercises 5 / 15
- 7. Images/cinvestav- Binary search trees: Operations This property allows to print the keys in sorted order! Inorder-tree-walk(x) 1 if x= NIL 2 Inorder-tree-walk(x.left) 3 print x.key 4 Inorder-tree-walk(x.right) 6 / 15
- 8. Images/cinvestav- Cost of inorder walk Theorem 12.1 If x is the root of an n-node subtree, then the call Inorder-tree-walk(x) takes Θ (n) time. Proof: Let T(n) denote the time taken by Inorder-tree-walk(x) when called at the root. First Since Inorder-tree-walk(x) visit all the nodes then we have that T (n) = Ω (n). Thus, you need to prove T (n) = O (n)? 7 / 15
- 9. Images/cinvestav- Cost of inorder walk Theorem 12.1 If x is the root of an n-node subtree, then the call Inorder-tree-walk(x) takes Θ (n) time. Proof: Let T(n) denote the time taken by Inorder-tree-walk(x) when called at the root. First Since Inorder-tree-walk(x) visit all the nodes then we have that T (n) = Ω (n). Thus, you need to prove T (n) = O (n)? 7 / 15
- 10. Images/cinvestav- Cost of inorder walk Theorem 12.1 If x is the root of an n-node subtree, then the call Inorder-tree-walk(x) takes Θ (n) time. Proof: Let T(n) denote the time taken by Inorder-tree-walk(x) when called at the root. First Since Inorder-tree-walk(x) visit all the nodes then we have that T (n) = Ω (n). Thus, you need to prove T (n) = O (n)? 7 / 15
- 11. Images/cinvestav- Proof of inorder walk First For n = 0, the method takes a constant time T(0) = c for some c > 0. Now for n > 0 We have the following situation: 1 Left subtree has k nodes 2 Right subtree has n − k − 1 nodes 8 / 15
- 12. Images/cinvestav- Proof of inorder walk First For n = 0, the method takes a constant time T(0) = c for some c > 0. Now for n > 0 We have the following situation: 1 Left subtree has k nodes 2 Right subtree has n − k − 1 nodes 8 / 15
- 13. Images/cinvestav- Substitution Method We have finally T(n) = T(k) + T(n − k − 1) + d 1 T(k) is the amount of work done in the left 2 T(n − k − 1) is the amount of work done in the right 3 d > 0 reflects an upper bound for the in-between work done for the print. We use the substitution method to prove that T(n) = O(n) This can be done if we can bound T(n) by bounding it by (c + d)n + c (1) Thus for n = 0 T (0) = c = (c + d) × 0 + c (2) 9 / 15
- 14. Images/cinvestav- Substitution Method We have finally T(n) = T(k) + T(n − k − 1) + d 1 T(k) is the amount of work done in the left 2 T(n − k − 1) is the amount of work done in the right 3 d > 0 reflects an upper bound for the in-between work done for the print. We use the substitution method to prove that T(n) = O(n) This can be done if we can bound T(n) by bounding it by (c + d)n + c (1) Thus for n = 0 T (0) = c = (c + d) × 0 + c (2) 9 / 15
- 15. Images/cinvestav- Substitution Method We have finally T(n) = T(k) + T(n − k − 1) + d 1 T(k) is the amount of work done in the left 2 T(n − k − 1) is the amount of work done in the right 3 d > 0 reflects an upper bound for the in-between work done for the print. We use the substitution method to prove that T(n) = O(n) This can be done if we can bound T(n) by bounding it by (c + d)n + c (1) Thus for n = 0 T (0) = c = (c + d) × 0 + c (2) 9 / 15
- 16. Images/cinvestav- Substitution Method For n > 0 T(n) ≤ T(k) + T(n − k − 1) + d = ((c + d) k + c) + ((c + d) (n − k − 1) + c) + d = (c + d) n + c − (c + d) + c + d = (c + d) n + c Thus T (n) = Θ (n) (3) 10 / 15
- 17. Images/cinvestav- Substitution Method For n > 0 T(n) ≤ T(k) + T(n − k − 1) + d = ((c + d) k + c) + ((c + d) (n − k − 1) + c) + d = (c + d) n + c − (c + d) + c + d = (c + d) n + c Thus T (n) = Θ (n) (3) 10 / 15
- 18. Images/cinvestav- Binary search trees: Operations Searching Tree-search(x, k) 1 if x == NIL or k == x.key 2 return x 3 if k < x.key 4 return Tree-search(x.left, k) 5 else return Tree-search(x.right, k) Complexity O (h) (4) where h is the height of the tree ⇒ we look for well balanced trees. 11 / 15
- 19. Images/cinvestav- Binary search trees: Operations Searching Tree-search(x, k) 1 if x == NIL or k == x.key 2 return x 3 if k < x.key 4 return Tree-search(x.left, k) 5 else return Tree-search(x.right, k) Complexity O (h) (4) where h is the height of the tree ⇒ we look for well balanced trees. 11 / 15
- 20. Images/cinvestav- Binary search trees: Operations Minimum and maximum Tree-minimum(x) 1 while x.left =NIL 2 x = x.left 3 return x Complexity O (h) (5) where h is the height of the tree ⇒ we look for well balanced trees. 12 / 15
- 21. Images/cinvestav- Binary search trees: Operations Minimum and maximum Tree-minimum(x) 1 while x.left =NIL 2 x = x.left 3 return x Complexity O (h) (5) where h is the height of the tree ⇒ we look for well balanced trees. 12 / 15
- 22. Images/cinvestav- Binary search trees: Operations TREE-DELETE(T, z) 1 if z.left == NIL 2 Transplant(T, z, z.right) 3 elseif z.right == NIL 4 Transplant(T, z, z.left) 5 else 6 y=Tree-minimum(z.right) 7 if y.p = z 8 Transplant(T, y, y.right) 9 y.right = z.right 10 y.right.p = y 11 Transplant(T, z, y) 12 y.left = z.left 13 y.left.p = y Case 1 Basically if the element z to be deleted has a NIL left child simply replace z with that child!!! 13 / 15
- 23. Images/cinvestav- Binary search trees: Operations TREE-DELETE(T, z) 1 if z.left == NIL 2 Transplant(T, z, z.right) 3 elseif z.right == NIL 4 Transplant(T, z, z.left) 5 else 6 y=Tree-minimum(z.right) 7 if y.p = z 8 Transplant(T, y, y.right) 9 y.right = z.right 10 y.right.p = y 11 Transplant(T, z, y) 12 y.left = z.left 13 y.left.p = y Case 2 Basically if the element z to be deleted has a NIL right child simply replace z with that child!!! 13 / 15
- 24. Images/cinvestav- Binary search trees: Operations TREE-DELETE(T, z) 1 if z.left == NIL 2 Transplant(T, z, z.right) 3 elseif z.right == NIL 4 Transplant(T, z, z.left) 5 else 6 y=Tree-minimum(z.right) 7 if y.p = z 8 Transplant(T, y, y.right) 9 y.right = z.right 10 y.right.p = y 11 Transplant(T, z, y) 12 y.left = z.left 13 y.left.p = y Case 3 The z element has not empty children you need to find the successor of it. 13 / 15
- 25. Images/cinvestav- Binary search trees: Operations TREE-DELETE(T, z) 1 if z.left == NIL 2 Transplant(T, z, z.right) 3 elseif z.right == NIL 4 Transplant(T, z, z.left) 5 else 6 y=Tree-minimum(z.right) 7 if y.p = z 8 Transplant(T, y, y.right) 9 y.right = z.right 10 y.right.p = y 11 Transplant(T, z, y) 12 y.left = z.left 13 y.left.p = y Case 4 if y.p = z then y.right takes the position of y after all y.left == NIL take z.right and make it the new right of y make the (y.right == z.right).p equal to y 13 / 15
- 26. Images/cinvestav- Binary search trees: Operations TREE-DELETE(T, z) 1 if z.left == NIL 2 Transplant(T, z, z.right) 3 elseif z.right == NIL 4 Transplant(T, z, z.left) 5 else 6 y=Tree-minimum(z.right) 7 if y.p = z 8 Transplant(T, y, y.right) 9 y.right = z.right 10 y.right.p = y 11 Transplant(T, z, y) 12 y.left = z.left 13 y.left.p = y Case 4 put y in the position of z make y.left equal to z.left make the (y.left == z.left).p equal to y 13 / 15
- 27. Images/cinvestav- Support Operations Transplant(T, u, v) 1 if u.p == NIL 2 T.root = v 3 elseif u == u.p.left 4 u.p.left = v 5 else u.p.right = v 6 if v =NIL 7 v.p = u.p Case 1 If u is the root then make the root equal to v 14 / 15
- 28. Images/cinvestav- Support Operations Transplant(T, u, v) 1 if u.p == NIL 2 T.root = v 3 elseif u == u.p.left 4 u.p.left = v 5 else u.p.right = v 6 if v =NIL 7 v.p = u.p Case 2 if u is the left child make the left child of the parent of u equal to v 14 / 15
- 29. Images/cinvestav- Support Operations Transplant(T, u, v) 1 if u.p == NIL 2 T.root = v 3 elseif u == u.p.left 4 u.p.left = v 5 else u.p.right = v 6 if v =NIL 7 v.p = u.p Case 3 Similar to the second case, but for right child 14 / 15
- 30. Images/cinvestav- Support Operations Transplant(T, u, v) 1 if u.p == NIL 2 T.root = v 3 elseif u == u.p.left 4 u.p.left = v 5 else u.p.right = v 6 if v =NIL 7 v.p = u.p Case 4 If v = NIL then make the parent of v the parent of u 14 / 15
- 31. Images/cinvestav- Excercises From Cormen’s book, chapters 11 and 12 11.1-2 11.2-1 11.2-2 11.2-3 11.3-1 11.3-3 12.1-3 12.1-5 12.2-5 12.2-7 12.2-9 12.3-3 15 / 15