This set of Data Structure Multiple Choice Questions & Answers (MCQs) focuses on “Preorder Traversal”.
1. For the tree below, write the pre-order traversal.
a) 2, 7, 2, 6, 5, 11, 5, 9, 4
b) 2, 7, 5, 2, 6, 9, 5, 11, 4
c) 2, 5, 11, 6, 7, 4, 9, 5, 2
d) 2, 7, 5, 6, 11, 2, 5, 4, 9
View Answer
Explanation: Pre order traversal follows NLR(Node-Left-Right).
2. For the tree below, write the post-order traversal.
a) 6, 2, 7, 2, 5, 11, 9, 5, 4
b) 6, 5, 11, 2, 7, 5, 9, 4, 2
c) 6, 5, 2, 11, 7, 4, 9, 5, 2
d) 6, 2, 7, 2, 11, 5, 5, 9, 4
View Answer
Explanation: Post order traversal follows LRN(Left-Right-Node).
3. Select the code snippet which performs pre-order traversal.
a)
public void preorder(Tree root) { System.out.println(root.data); preorder(root.left); preorder(root.right); }
b)
public void preorder(Tree root) { preorder(root.left); System.out.println(root.data); preorder(root.right); }
c)
public void preorder(Tree root) { System.out.println(root.data); preorder(root.right); preorder(root.left); }
d)
public void preorder(Tree root) { preorder(root.right); preorder(root.left); System.out.println(root.data); }
Explanation: Pre-order traversal follows NLR(Node-Left-Right).
4. Select the code snippet which performs post-order traversal.
a)
public void postorder(Tree root) { System.out.println(root.data); postorder(root.left); postorder(root.right); }
b)
public void postorder(Tree root) { postorder(root.left); postorder(root.right); System.out.println(root.data); }
c)
public void postorder(Tree root) { System.out.println(root.data); postorder(root.right); postorder(root.left); }
d)
public void postorder(Tree root) { postorder(root.right); System.out.println(root.data); postorder(root.left); }
Explanation: Post order traversal follows NLR(Left-Right-Node).
5. Select the code snippet that performs pre-order traversal iteratively.
a)
public void preOrder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); st.add(root); while (!stk.empty()) { Tree node = stk.pop(); System.out.print(node.data + " "); if (node.left != null) stk.push(node.left); if (node.right != null) stk.push(node.right); } }
b)
public void preOrder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); while (!stk.empty()) { Tree node = stk.pop(); System.out.print(node.data + " "); if (node.right != null) stk.push(node.right); if (node.left != null) stk.push(node.left); } }
c)
public void preOrder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); st.add(root); while (!stk.empty()) { Tree node = stk.pop(); System.out.print(node.data + " "); if (node.right != null) stk.push(node.right); if (node.left != null) stk.push(node.left); } }
d)
public void preOrder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); st.add(root); while (!stk.empty()) { Tree node = stk.pop(); System.out.print(node.data + " "); if (node.right != null) stk.push(node.left); if (node.left != null) stk.push(node.right); } }
Explanation: Add the root to the stack first, then continously check for the right and left children of the top-of-the-stack.
6. Select the code snippet that performs post-order traversal iteratively.
a)
public void postorder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); Tree node = root; while (!stk.isEmpty() || node != null) { while (node != null) { if (node.right != null) stk.add(node.left); stk.add(node); node = node.right; } node = stk.pop(); if (node.right != null && !stk.isEmpty() && node.right == stk.peek()) { Trees nodeRight = stk.pop(); stk.push(node); node = nodeRight; } else { System.out.print(node.data + " "); node = null; } } }
b)
public void postorder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); Tree node = root; while (!stk.isEmpty() || node != null) { while (node != null) { if (node.right != null) stk.add(node.right); stk.add(node); node = node.left; } node = stk.pop(); if (node.right != null && !stk.isEmpty() && node.right == stk.peek()) { Trees nodeRight = stk.pop(); stk.push(node); node = nodeRight; } else { System.out.print(node.data + " "); node = null; } } }
c)
public void postorder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); Tree node = root; while (!stk.isEmpty() || node != null) { while (node != null) { if (node.right != null) stk.add(node.right); stk.add(node); node = node.left; } node = stk.pop(); if (node.right != null) { Trees nodeRight = stk.pop(); stk.push(node); node = nodeRight; } else { System.out.print(node.data + " "); node = null; } } }
d)
public void postorder(Tree root) { if (root == null) return; Stack<Tree> stk = new Stack<Tree>(); Tree node = root; while (!stk.isEmpty() || node != null) { while (node != null) { if (node.right != null) stk.add(node.left); stk.add(node); node = node.left; } node = stk.pop(); if (node.right != null) { Trees nodeRight = stk.pop(); stk.push(node); node = nodeLeft; } else { System.out.print(node.data + " "); node = null; } } }
Explanation: The left and right children are added first to the stack, followed by the node, which is then popped. Post-order follows LRN policy.
7. What is the time complexity of pre-order traversal in the iterative fashion?
a) O(1)
b) O(n)
c) O(logn)
d) O(nlogn)
View Answer
Explanation: Since you have to go through all the nodes, the complexity becomes O(n).
8. What is the space complexity of the post-order traversal in the recursive fashion? (d is the tree depth and n is the number of nodes)
a) O(1)
b) O(nlogd)
c) O(logd)
d) O(d)
View Answer
Explanation: In the worst case we have d stack frames in the recursive call, hence the complexity is O(d).
9. To obtain a prefix expression, which of the tree traversals is used?
a) Level-order traversal
b) Pre-order traversal
c) Post-order traversal
d) In-order traversal
View Answer
Explanation: As the name itself suggests, pre-order traversal can be used.
10. Consider the following data. The pre order traversal of a binary tree is A, B, E, C, D. The in order traversal of the same binary tree is B, E, A, D, C. The level order sequence for the binary tree is _________
a) A, C, D, B, E
b) A, B, C, D, E
c) A, B, C, E, D
d) D, B, E, A, C
View Answer
Explanation: The inorder sequence is B, E, A, D, C and Preorder sequence is A, B, E, C, D. The tree constructed with the inorder and preorder sequence is
The levelorder traversal (BFS traversal) is A, B, C, E, D.
11. Consider the following data and specify which one is Preorder Traversal Sequence, Inorder and Postorder sequences.
S1: N, M, P, O, Q
S2: N, P, Q, O, M
S3: M, N, O, P, Q
a) S1 is preorder, S2 is inorder and S3 is postorder
b) S1 is inorder, S2 is preorder and S3 is postorder
c) S1 is inorder, S2 is postorder and S3 is preorder
d) S1 is postorder, S2 is inorder and S3 is preorder
View Answer
Explanation: Preorder traversal starts from the root node and postorder and inorder starts from the left child node of the left subtree. The first node of S3 is different and for S1 and S2 it’s the same. Thus, S3 is preorder traversal and the root node is M. Postorder traversal visits the root node at last. S2 has the root node(M) at last that implies S2 is postorder traversal. S1 is inorder traversal as S2 is postorder traversal and S3 is preorder traversal. Therefore, S1 is inorder traversal, S2 is postorder traversal and S3 is preorder traversal.
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