Ne=N2+1
Here Ne=no. of leaf nodes
N2= no. of nodes of degree 2
when it is a hundred stem and leaf plot you would put the first two numbers of that number in the stem part; and the last number in the leaf part and you would continue on as if it were a two digit number.
If the leaf stem is present, it has symmetry of order 1. Otherwise, it can have rotational symmetry of order 3.
The stem is the tens place, and the leaf is the ones. The leaf has a few choices. It's used to place a number.
Stem|Leaf 3|0 where the number of tens goes on the left and the number of ones (<10) goes on the right.
mode is the frequent number appeared in the number plot.
The height of a complete binary tree is in terms of log(n) where n is the number of nodes in the tree. The height of a complete binary tree is the maximum number of edges from the root to a leaf, and in a complete binary tree, the number of leaf nodes is equal to the number of internal nodes plus 1. Since the number of leaf nodes in a complete binary tree is equal to 2^h where h is the height of the tree, we can use log2 to find the height of a complete binary tree in terms of the number of nodes.
Complete Binary tree: -All leaf nodes are found at the tree depth level -All nodes(non-leaf) have two children Strictly Binary tree: -Nodes can have 0 or 2 children
Complete Binary tree: All leaf nodes are found at the tree depth level and All non-leaf nodes have two children. Extended Binary tree: Nodes can have either 0 or 2 children.
A strictly binary tree is a tree in which every node other than the leaf nodes has exactly two children. OR in the Graph Theory perspective a tree having it's root vertex with degree 2 and all other non-leaf vertex of degree 3 and leaf vertex of degree 1, is called as the strictly binary tree. it is also called as the 2-tree or full binary tree.
IF EVERY NON-LEAF NODE IN A BINARY TREE HAS HAS NONEMPTY LEFT AND RIGHT SUBTREES, THE TREE IS TERMED AS A STRICTLY BINARY TREE. SUCH A TREE WITH n LEAVES ALWAYS CONTAINS 2n-1 NODES.
For a full binary tree of height 3 there are 4 leaf nodes. E.g., 1 root, 2 children and 4 grandchildren.
4
It will be come a terminal node. Normally we call terminal nodes leaf nodes because a leaf has no branches other than its parent.
No. A leaf node is a node that has no child nodes. A null node is a node pointer that points to the null address (address zero). Since a leaf node has no children, its child nodes are null nodes.
Convert n to a binary value, then set the next most significant bit. For instance, if there are 7 non-leaves, this equates to 00000111 in binary. Each bit tells us how many non-leaves exist in each level, where the least-significant bit (bit 0) represents the root node and the most-significant bit (bit 2) represents the lowest level. Thus we have 1+2+4=7 non-leaf nodes in total. The next most-significant bit (bit 3) represents the leaf nodes and if we set that bit we get 00001111, which is 15 decimal. Thus there are 15 nodes in total. We can visualise this binary tree using hexadecimal notation: 1 2 3 4 5 6 7 8 9 a b c d e f (Note: 0xf = 15 decimal). Using binary notation, we get the following: 1st level (bit 0) = 00000001 = 1 non-leaf node (the root) 2nd level (bit 1) = 00000010 = 2 non-leaf nodes 3rd level (bit 2) = 00000100 = 4 non-leaf nodes 4th level (bit 3) = 00001000 = 8 leaf nodes Thus we get: 00000001+00000010+00000100+00001000=00001111 Or, in decimal: 1+2+4+8=15
Let N = the number of nodes, F = number of full nodes, L = the number of leaves, and H = the number of nodes with one child (or half nodes). The total number of nodes in a binary tree equals N = F + H + L. Because each full node is incident on two outgoing edges, each half node is incident on one outgoing edge, and each leaf is incident on no outgoing edge it follows that the total number of edges in a binary tree equals 2F + H. It is also true that the total number of edges in a tree equals N 1. Thus, 2F + H = N 1 H = N 1 2F Subbing this value of H into N = F + H + L gives, N = F + N 1 2F + L N N + 1 = F + L F + 1 = L
int countleaves(struct node* root){ if(root!=null) { countleaves(root->left); if(root->left==NULL&&root->right==NULL) { count++; } countleaves(root->right); } }