Hint: Assume the binary tree is properly balanced (the depth of the tree T is O(log N)). For full credit, your algorithm should run in O(K+log N) average time, where K is the number of keys printed. Solution // Since the language is not specified, I\'m doing it in C++. // Assumption: structure of the tree node: /* struct Node{ int data; Node *left; Node *right; }; */ void method(Node *root, int k1, int k2){ if(root == NULL) return; method(root->left); if(root->data >= k1 && root->data <= k2) cout << root->data << \" \"; method(root->right); }.