20 + 21 +... 2h = 2(h+1)-1 will be the total number of nodes. For example, the binary tree with height 2 illustrated in Figure 2(b) contains 2(2+1)-1 = 7 nodes. The left child of a node in a binary search tree has a value less than the parent, while the right child has a value larger than the parent. In the example shown, the first cell on the grid is one bit, and it's set to 1, which means that the root is a left child and its value is less than 1. The second cell from the left is another bit and it's set to 0, which means that the root is the only child and its value is equal to 0. Continuing in **this way**, we can see that the last cell on the grid is the seventh cell from the left and it has a value of 1, which means that there are six cells with values of 1 next to it and one empty cell that should have a value of 0 but doesn't because it's the only cell on the grid.

- How do you count the number of nodes in a binary tree?
- How do you find the maximum number of nodes in a tree?
- How many minimum nodes are needed to create a tree?
- What is the maximum number of elements at height h in a binary tree whose root is at height 0?
- Which expression gives the maximum number of nodes?
- How many leaf nodes are present in a binary tree having a depth of H?
- How many sons are there in a binary tree?
- What is the minimum and maximum number of nodes possible on a level 5 binary search tree?

The smallest number of nodes in a binary search tree with a height of h is h+1 (in the case of left-skewed and right-skewed binary search trees). When all levels of a binary search tree are entirely full, the maximum number of nodes is h. There will be a total of **20 + 21 +**... nodes. 2h is equal to 2(h+1)-1. Therefore, the maximum number of nodes in a binary search tree is 2(h+1)-1.

Each of the first two h levels must have at least one node. At the first h layers, all potential nodes are present. A complete binary tree with h nodes contains 2h minus 1 nodes. 4 entire binary trees in height.

2 A binary tree of height 'h' can have a maximum of 2h-1 nodes. In this context, a tree's height is defined as the number of nodes on its root-to-leaf route. The height of a tree with **a single node** is taken to be 1. Thus, the maximum number of elements at height 'h' in a binary tree whose root is at height 0 is 2^(h-1). Using this formula and a computer search we can verify that the above example tree has a maximum size of 13.

At level 'l,' the maximum number of nodes is 2l-1. The number of nodes on the path from root to node, including the root itself, is denoted as level. We are assuming that the root level is 1. A binary tree of **height h** can have a maximum of 2h-1 nodes.

In general, a binary tree with n levels will contain **2^n-1 nodes**.

Note that for a binary tree, the number of nodes will always be a perfect square, because if it was not a square, then there would be a number of nodes in the tree that could not be reached. For example, if the tree contained 3 levels, then it would contain 9 nodes: 2^3-1 = 8, 7, 6, 5, 4, 3, 2, 1. A tree with **11 nodes** in it cannot exist; there would be no way to reach the highest level.

For our special case where l is a positive integer, the maximum number of nodes at **any level** is 2l-1. There can be at most two layers, the first having length l and the second length l-1, so the total number of nodes is at most 2l-1 + 2l-1 = 5l-2. This bound is achieved by a tree with three layers, each containing lengths of 1, 2, and 3 elements, respectively.

There are two leaf nodes in a perfect binary tree of **height h.** For the height, we'll utilize induction once more. There are 20 nodes when h = 0, and one of them is a leaf node. 2*2h = 2h + 1 leaf node is required for h + 1.

The number of leaves in a binary tree is n+1 since each non-leaf node has precisely two sons. There are 2n+1 nodes in all. A complete binary tree is one that contains two offspring for every node except the leaves. There are exactly such trees that contain n distinct objects.

The left-skewed binary tree in Figure 1 (a) with 5 nodes has a height of 5-1 = 4, but the binary tree in Figure 1 (b) with 5 nodes has a height of floor (log25) = 2. The smallest number of nodes in a binary tree with a height of h is h+1 (in the case of left-skewed and right-skewed binary trees). Therefore, the minimum number of nodes in a level-5 binary search tree is 6.

The maximum number of nodes in a level-5 binary search tree is not clear because there can be any number of nodes at each level. However, we can assume that the number of nodes at each level is close to **an average value** by considering the numbers of nodes at **different levels** in randomly generated binary search trees of the same size. FIGURE 1: Examples of left-skewed and right-skewed binary search trees with 5 nodes at each level.

In random binary search trees of size n, the number of nodes at **each level** is approximately equal to n/2, so the average number of nodes at each level is n/2. Thus, the maximum number of nodes in a level-5 binary search tree is ceil(n/2)*6 = n*3/2 - 3.

Note that the number of nodes in a level-5 binary search tree does not necessarily have to be an integer. For example, a tree with 7 nodes at **each level** is also a valid level-5 binary search tree.