Question
Formatted question description: https://leetcode.ca/all/173.html
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.
Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
@tag-tree
Algorithm
The non-recursive form of the in-order traversal of the binary tree requires an additional definition of a stack to assist. The building rule of the binary search tree is left<root<right
, and the in-order traversal can extract all nodes from small to large.
Code
Java
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Stack;
public class Binary_Search_Tree_Iterator {
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
Stack<TreeNode> sk;
// @note:@memorize: important feature, min is always going left branch to leaf
public BSTIterator(TreeNode root) {
sk = new Stack<>();
// all the way to leftmost leaf
while (root != null) {
sk.push(root);
root = root.left;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !sk.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode minNode = sk.pop();
TreeNode current = minNode;
// update stack, possible next time min is frmo its right-then-left branch
if (current.right != null) {
current = current.right;
// same logic as in constructor
while (current != null) {
sk.push(current.left);
current = current.left;
}
}
return minNode.val;
}
}
public class BSTIteratorrrrrr {
List<Integer> sorted;
public BSTIteratorrrrrr(TreeNode root) {
sorted = new ArrayList<>();
dfs(root);
Collections.sort(sorted); // NlogN
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
sorted.add(root.val);
dfs(root.left);
dfs(root.right);
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !sorted.isEmpty();
}
/** @return the next smallest number */
public int next() {
return sorted.remove(0);
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
}