# Recent questions tagged binary-tree

1
A complete $n$-ary tree is a tree in which each node has $n$ children or no children. Let $I$ be the number of internal nodes and $L$ be the number of leaves in a complete $n$-ary tree. If $L=41$, and $I=10$, what is the value of $n$? $3$ $4$ $5$ $6$
2
A full binary tree with $n$ non-leaf nodes contains $\log_ 2 n$ nodes $n+1$ nodes $2n$ nodes $2n+1$ nodes
3
The height of a binary tree is the maximum number of edges in any root to leaf path. The maximum number number of nodes in a binary tree of height $h$ is $2^{h}$ $2^{h-1} – 1$ $2^{h+1} – 1$ $2^{h+1}$
4
The number of possible binary trees with $4$ nodes is $12$ $13$ $14$ $15$
5
In a full binary tree number of nodes is $63$ then the height of the tree is : $2$ $4$ $3$ $6$
6
Traversing a binary tree first root and then left and right subtrees called ______ traversal. postorder. preorder. inorder. none of these.
7
The number of unused pointers in a complete binary tree of depth $5$ is: $4$ $8$ $16$ $32$
8
Which of the following need not be a binary tree? Search tree Heap AVL tree B tree
9
The maximum number of nodes in a binary tree of level $k, k\geq1$ is: $2^k+1$ $2^{k-1}$ $2^k-1$ $2^{k-1}-1$
10
Consider a complete binary tree where the left and the right sub trees of the root are max-heaps. The lower bound for the number of operations to convert the tree to a heap is: $\Omega(\log n)$ $\Omega(n\log n)$ $\Omega(n)$ $\Omega(n^2)$
11
Which of the following statements is false? Optimal binary search tree construction can be performed efficiently using dynamic programming. Breadth-first search cannot be used to find connected components of a graph. Given the prefix and postfix walks of a binary tree, the tree cannot be reconstructed uniquely. Depth-first-search can be used to find the connected components of a graph. a b c d
12
The post-order traversal of binary tree is $ACEDBHIGF$. The pre-order traversal is $\text{A B C D E F G H I}$ $\text{F B A D C E G I H}$ $\text{F A B C D E G H I}$ $\text{A B D C E F G I H}$
1 vote
13
Argue that since sorting $n$ elements takes $\Omega (n\ lgn)$ time in the worst case in the comparison model, any comparison-based algorithm for constructing a $BST$ from an arbitrary list of n elements takes $\Omega (n\ lgn)$ time in the worst case.
1 vote
14
Consider the following function height, to which pointer to the root node of a binary tree shown below is passed Note that max(a,b) defined by #define max(a,b) (a>b)?a:b. int height(Node *root) The output of the above code will be _________________
1 vote
15
The number of node in each left subtree is within a factor of $2.$ of the number of nodes in the corresponding right subtree. Also a node allowed to have only one child if that child has no children. This tree has worst case height $O(logn)$. $N$ is the number of nodes in the binary tree. Is this statement TRUE about Binary Tree?
16
What is the time complexity for insertion in binary tree in worst case? O(1) O(log n) O(n) O(n log n)
17
somewhere we seen that formula- How many binary tree possible without labeled =c(2n,n)/n+1. anybody explain how we get this formula.
18
Let $T$ be a full binary tree with $8$ leaves. (A full binary tree has every level full.) Suppose two leaves $a$ and $b$ of $T$ are chosen uniformly and independently at random. The expected value of the distance between $a$ and $b$ in $T$ (ie., the number of edges in the unique path between $a$ and $b$) is (rounded off to $2$ decimal places) _________.
19
Which of the following is not correct about B + Tree, which is used for creating index of relational database table? (a) Key values in each node kept in sorted order (b) Leaf node pointer points to next node (c) B + tree is height balanced tree (d) Non-leaf node have pointers to data records
20
The height of a binary tree is defined as the number of nodes in the longest path from root to the leaf node. Let X be the height of a complete binary tree with 256 nodes. Then the value of X will be Answer 9
21
I think its answer is 8 .Please ,can any one make it sure for me :)
22
Consider a binary tree for every node | P - Q | <= 2. P represents number of nodes in left subtree of S and Q represents number of nodes in right subtree of S for h > 0. The minimum number of nodes present in such tree of height h = 4 ( Root at 0 level)
Let us there are n nodes which are labelled. Then the number of trees possible is given by the Catalan Number i.e $\binom{2n}{n} / (n+1)$ Then the binary search trees possible is just 1?