LeetCode 173 Binary Search Tree Iterator

LeetCode 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.

题目大意：

实现BST的Iterator

算法思路：

参照KB中BST的非递归中序遍历。将其分拆为初始化以及去掉stack不为空的循环分别为所求。

注意事项：

1. 初始化，将root所有左节点加入到stack。先写next，出栈栈顶节点，将它的右儿子的所有左节点入栈。

Python代码：

class BSTIterator(TestCases):

    def __init__(self, root: TreeNode):
        self.stack = []
        it = root
        while it:
            self.stack.append(it)
            it = it.left

    def next(self) -> int:
        node = self.stack.pop()
        if node.right:
            n = node.right
            while n:
                self.stack.append(n)
                n = n.left
        return node.val

    def hasNext(self) -> bool:
        return True if self.stack else False

Java代码：

Stack<TreeNode> s = new Stack<>();
public L173BinarySearchTreeIterator(TreeNode root) {
	TreeNode head = root;
	while(head != null) {
		s.push(head);
		head = head.left;
	}
}

// Recommended
public int next2() {
	TreeNode node = s.pop(); 
	if(node.right != null) {// left node has been visited
		TreeNode n = node.right; 
		while(n != null) {
			s.push(n);
			n = n.left;
		}
	}
	return node.val;

}

/** @return whether we have a next smallest number */
public boolean hasNext() {
	return !s.isEmpty();
}

算法分析：

next的平均时间复杂度(amortized complexity)为O(1)，n为字符串长度，空间复杂度O(logn)。