ii The height or depth of a binary tree is the maxi mum depth of any node or 1 if the tree is empty Any binary tree can have at most 2 nodes at depth Easy proof by induction EFINITION A complete binary tree of height is a binary tree which contain ID: 19553
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Examples: Not nearly complete: ( a ) Not nearly complete: ( b ) fails. Nearly complete. 1 2 3 4 5 6 7 8 9 1011121314151617 18 19202122J Q D B T Z G LCUANFRHVI E WKMY , and 5/25+1 == 11, and we find 4567891012131115 42 1 2 3 4 5 6 7 8 9 1011121314151617 18 19202122 81 Heap condition at node p 1)/2 pr In general, the worst-case number of comparisons to lg(maxlg( lg(max(lg( represented in an array A of size at least n+ ( A, n = n + 1; A[ n�] A[ ¬n/2¼] ) n], A[ ¬n/2¼] ); Find the largest element Given a nearly complete binary tree, in which the heap property can fail only at the children of the root, we can make the tree into a heap using a procedure called max-heapify() Among the root and its two children (nodes qwe find the largest element. (Two comparisons In this case, the largest (72) occurs in node r If the largest of these three elements were to occur in the root (not the case here), we would be done. If the largest occurs in a child of the root (as happens here, node r), we exchange the element in the root with the element in this child. 29 heap propertycould fail here q rs With the array representation, we can write max-heapify() like this. // A is an array of size at least n, which we think of // as a nearly complete binary tree. In the subtree // of A[1..n] rooted at A[i] , the heap property // holds everywhere except possibly at the children // of A[i]. This function makes the subtree of // A[1..n] rooted at A[i] into a heap. ( A, = i A[2 largest = 2 +1 A[2+1]� A[ +1; i ) i], A[largest]) ( A, We can also write max-heapify() non-recursively like this: However, we can not , decide if the heap contains an element x , find the k largest element in the heap (unless is 1, or at least is very close to 1). , remove x from the heap, if it is present.