1) The maximum number of nodes at level ‘l’ of a binary tree is 2l-1.
Here level is number of nodes on path from root to the node (including root and node). Level of root is 1.
This can be proved by induction.
For root, l = 1, number of nodes = 21-1 = 1
Assume that maximum number of nodes on level l is 2l-1
Since in Binary tree every node has at most 2 children, next level would have twice nodes, i.e. 2 * 2l-1
2) Maximum number of nodes in a binary tree of height ‘h’ is 2h – 1.
Here height of a tree is maximum number of nodes on root to leaf path. Height of a leaf node is considered as 1.
This result can be derived from point 2 above. A tree has maximum nodes if all levels have maximum nodes. So maximum number of nodes in a binary tree of height h is 1 + 2 + 4 + .. + 2h-1. This is a simple geometric series with h terms and sum of this series is 2h – 1.
In some books, height of a leaf is considered as 0. In this convention, the above formula becomes 2h+1 – 1
3) In a Binary Tree with N nodes, minimum possible height or minimum number of levels is ⌈ Log2(N+1) ⌉
This can be directly derived from point 2 above. If we consider the convention where height of a leaf node is considered as 0, then above formula for minimum possible height becomes ⌈ Log2(N+1) ⌉ – 1
4) A Binary Tree with L leaves has at least ⌈ Log2L ⌉ + 1 levels
A Binary tree has maximum number of leaves when all levels are fully filled. Let all leaves be at level l, then below is true for number of leaves L.
L <= 2l-1 [From Point 1]
Log2L = ⌈ Log2L ⌉ + 1
5) In Binary tree, number of leaf nodes is always one more than nodes with two children.
L = T + 1
Where L = Number of leaf nodes
T = Number of internal nodes with two children
See Handshaking Lemma and Tree for proof.
In the next article on tree series, we will be discussing different types of Binary Trees and their properties.
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