173. Binary Search Tree Iterator

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:

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.

Solution

非递归中序遍历

Complexity

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class BSTIterator {

    private Deque<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new ArrayDeque<>();
        while(root!=null){
            stack.push(root);
            root = root.left;
        }
    }
    
    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.pop();
        TreeNode right = node.right;
        while(right != null){
            stack.push(right);
            right = right.left;
        }
        return node.val;
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return stack.size() > 0;
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */