173 Binary Search Tree Iterator

173. Binary Search Tree Iterator

1. Question

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

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

Note:next()andhasNext()should run in average O(1) time and uses O(h) memory, wherehis the height of the tree.

2. Implementation

思路: 其实就是Inorder Traversal分解成next() 和 hasNext()两个function

public class BSTIterator {
    TreeNode curNode = null;
    Stack<TreeNode> stack = null;

    public BSTIterator(TreeNode root) {
        curNode = root;
        stack = new Stack<>();
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty() || curNode != null;
    }

    /** @return the next smallest number */
    public int next() {
        while (curNode != null) {
            stack.push(curNode);
            curNode = curNode.left;
        }
        TreeNode temp = stack.pop();
        curNode = temp.right;
        return temp.val;
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

3. Time & Space Complexity

hasNext()时间复杂度是O(1), next()时间复杂度是O(n), 平均时间复杂度是O(1), 空间复杂度是O(h)