Question
Implement a binary tree with a method getRandomNode() that returns a random node.
eg.
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getRandomNode() = 5 getRandomNode() = 8 getRandomNode() = 1 |
Once you think that you’ve solved the problem, click below to see the solution.
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Solution
How was that problem? You can check out the solution in the video below.
Here is the source code for the solution shown in the video (Github):
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// Individual node of the tree private class Node { Node left; Node right; int val; int children; } // The root of the tree private Node root; private Random rand; // Return each node with probability 1/N public int getRandomNode() { if (root == null) throw new NullPointerException(); // This is an index of a node in the tree. Indices go in sorted order. int count = rand.nextInt(root.children + 1); return getRandomNode(root, count); } // Recursive method. Binary search through tree to find the index. We use // the number of children to determine which direction to go private int getRandomNode(Node curr, int count) { if (count == children(curr.left)) return curr.val; if (count < children(curr.left)) return getRandomNode(curr.left, count); // The new index becomes the index of the same node but now within the // subtree rather than the whole tree return getRandomNode(curr.right, count - children(curr.left) - 1); } // Return the number of nodes in a given subtree private int children(Node n) { if (n == null) return 0; return n.children + 1; } |
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