算法练习Stack(六)--P173

2020-03-07

栈stack
173. Binary Search Tree Iterator Medium

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.

这块的思路还没理清楚。跟二叉树的属性有关系。

private final Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        TreeNode cur = root;
        while (cur != null) {
            stack.push(cur);
            cur = cur.left;
        }
    }

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

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

        // Traversal cur node's right branch
        TreeNode cur = node.right;
        while (cur != null){
            stack.push(cur);
            cur = cur.left;
        }

        return node.val;
    }