[LeetCode] Problem 173 - Binary Search Tree Iterator

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Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Example

E3yZKf.png

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BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

Note

  • next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
  • You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

Code

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public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
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private Stack<TreeNode> stack = new Stack<>();

public BSTIterator(TreeNode root) {
    while (root != null) {
        stack.push(root);
        root = root.left;
    }
}

public int next() {
    TreeNode node = stack.pop();
    int val = node.val;

    if (node.right != null) {
        node = node.right;

        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    return val;
}

public boolean hasNext() {
    return !stack.isEmpty();
}