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A binary tree $T$ is constructed such that for every node $N$, the number of nodes in its left subtree $L(N)$ and right subtree $R(N)$ satisfy the condition: $\mid \operatorname{size}(L(N))- \operatorname{size}(R(N)) \mid \leq 1$.

If the total number of nodes in the tree is $\mathbf{1 5}$, what is the maximum possible height of the tree? (Assume a tree with a single node has a height of $0 )$.

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Step 1: Understand the conditions

The total number of nodes is $N = 15$.

The condition for every node $M$ is that the sizes of its left and right subtrees, $size(L(M))$ and $size(R(M))$, must satisfy the inequality $|size(L(M)) - size(R(M))| \le 1$.

 

Step 2: Determine subtree sizes

 

For the root node, we have $size(L(N)) + size(R(N)) + 1 = 15$, which simplifies to $size(L(N)) + size(R(N)) = 14$.

Given the condition $|size(L(N)) - size(R(N))| \le 1$, the only possible integer solution is $size(L(N)) = 7$ and $size(R(N)) = 7$.

This logic applies recursively to all subtrees, forcing the tree to be a perfect binary tree.

 

Step 3: Calculate the height

The number of nodes $N$ in a perfect binary tree of height $h$ (where a single node has height 0) is given by the formula $N = 2^{h+1} - 1$.

 

Step 4: Solve for the height

We substitute the total number of nodes $N = 15$ into the formula and solve for $h$:

$15 = 2^{h+1} - 1$

$16 = 2^{h+1}$

$2^4 = 2^{h+1}$

$h+1 = 4$

$h = 3$

 

Answer:   The maximum possible height of the tree under the given conditions is 3.
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