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Node Average Depth in a Full Binary Tree Timothy Rolfe In a completely full binary tree, the number of nodes (n) for a tree with height k is given by n = 2k+1 1. This defines n and k in the following discussion. To average the node depth, we simply need to total up the levels of each node in the binary search tree. At a given level j, there are 2 j tree nodes (since we number the root as level 0). Thus we simply need to total up, for each level, the product of the level with the number of nodes at that level. (We can, of course, omit the j=0 term.) The average will then be the total of these divided by the number of nodes in the tree. k j j 2 Average = j= 1 n Brian Carlson, formerly of Dakota State University in South Dakota, has discovered a closed-form solution for a closely related summation that can be rearranged into exactly this summation: j (1) 2 = ( k -1) 2 j j =1 k k +1 +2 *1 (2) We have already defined n as (2k+1 1). Combining equations (1) and (2): Average = (k -1) 2 + 2 As k grows, this simplifies to ( k +1 ) (2 k +1 -1 ) (3) Average (k -1) (4) In other words, as the tree grows, the maximum node depth (tree height) is just one more than the average node depth. [Rework of material in Timothy Rolfe's article "Analytic Derivation of Comparisons in Binary Search", SIGNUM Newsletter: Special Interest Group on Numerical Mathematics, Vol. 32, No. 4 (Oct 1997), pp. 1* S(k) = (k-1) x 2k+1 + 2 S(1) = (1 x 2 ) = (1-1) x + 2 = 2 S(k+1) = S(k) + (k+1) x 2k+1 = (k-1) x 2k+1 + 2 + (k+1) x 2k+1 = (2k) x 2k+1 + 2 = (k) x 2k+2 + 2 1 21 (Inductive hypothesis) (Base case confirmed) (Definition of recurrence) (Apply inductive hypo.) (Algebraic rearrangement) QED 15-19.]

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