Computers and Technology, 05.05.2020 05:28, klogsdon4380

1. Consider the following mystery function on a binary tree (not necessarily a BST). public int mysteryHelper(TreeNode cur){ if(cur==null) return 0; else{ int r=mysteryHelper(cur. getLeft()) + mysteryHelper(cur. getRight()); if(((cur. getLeft()!=null && ((Integer) cur. getLeft().getValue())pareTo(cur. getValue())>0)|| (cur. getRight()!=null && ((Integer) cur. getRight().getValue())pareTo(cur. getValue())<0)){ r++; } return r; } } public int mystery(){ return mysteryHelper(root); }

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1. Consider the following mystery function on a binary tree (not necessarily a BST). public int myst...