Tree Traversals - GeeksforGeeks
Tree Traversals Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways. Following are the generally used ways for traversing trees. Example Tree Please see this post for Breadth First Traversal. Inorder Traversal: Algorithm Inorder(tree) 1. Traverse the left subtree, i.e., call Inorder(left-subtree) 2. Visit the root. 3. Traverse the right subtree, i.e., call Inorder(right-subtree) Uses of Inorder In case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder itraversal s reversed, can be used. Example: Inorder traversal for the above given figure is 4 2 5 1 3. Preorder Traversal: Algorithm Preorder(tree) 1. Visit the root. 2. Traverse the left subtree, i.e., call Preorder(left-subtree) 3. Traverse the right subtree, i.e.,
Tree Traversals Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways. Following are the generally used ways for traversing trees. Example Tree Please see this post for Breadth First Traversal. Inorder Traversal: Algorithm Inorder(tree) 1. Traverse the left subtree, i.e., call Inorder(left-subtree) 2. Visit the root. 3. Traverse the right subtree, i.e., call Inorder(right-subtree) Uses of Inorder In case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder itraversal s reversed, can be used. Example: Inorder traversal for the above given figure is 4 2 5 1 3. Preorder Traversal: Algorithm Preorder(tree) 1. Visit the root. 2. Traverse the left subtree, i.e., call Preorder(left-subtree) 3. Traverse the right subtree, i.e.,
Time Complexity: O(n)
Let us prove it:
Let us prove it:
Complexity function T(n) — for all problem where tree traversal is involved — can be defined as:
T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of root and n-k-1 on the other side.
Let’s do analysis of boundary conditions
Case 1: Skewed tree (One of the subtrees is empty and other subtree is non-empty )
k is 0 in this case.
T(n) = T(0) + T(n-1) + c
T(n) = 2T(0) + T(n-2) + 2c
T(n) = 3T(0) + T(n-3) + 3c
T(n) = 4T(0) + T(n-4) + 4c
T(n) = T(0) + T(n-1) + c
T(n) = 2T(0) + T(n-2) + 2c
T(n) = 3T(0) + T(n-3) + 3c
T(n) = 4T(0) + T(n-4) + 4c
…………………………………………
………………………………………….
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0) + (n)c
………………………………………….
T(n) = (n-1)T(0) + T(1) + (n-1)c
T(n) = nT(0) + (n)c
Value of T(0) will be some constant say d. (traversing a empty tree will take some constants time)
T(n) = n(c+d)
T(n) = (-)(n) (Theta of n)
T(n) = (-)(n) (Theta of n)
Case 2: Both left and right subtrees have equal number of nodes.
T(n) = 2T(|_n/2_|) + c
This recursive function is in the standard form (T(n) = aT(n/b) + (-)(n) ) for master methodhttp://en.wikipedia.org/wiki/Master_theorem. If we solve it by master method we get (-)(n)
Auxiliary Space : If we don’t consider size of stack for function calls then O(1) otherwise O(n).
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