Binary Search Tree Iterator java - binary-search-tree

Question

I did the LeetCode question Binary Search Tree Iterator. the following code is what I learned from others. One part I didn't understand which is cur = cur.right. Since I got the smallest value of cur.val. Why do I need to assign cur.right to cur? When I remove cur = cur.right, it said time limit exceeded. Could someone can help me to explain it?
public class BSTIterator {
Stack<TreeNode> stack;
TreeNode cur;
public BSTIterator(TreeNode root) {
stack = new Stack<>();
cur = root;
}
/** #return the next smallest number */
public int next() {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
int val = cur.val;
cur = cur.right; //why needed to assign cur.right to cur?
return val;
}
}

Answer 1

By thinking on the structure of a binary search tree we know that the left node is less than the parent (or middle) node, and that the right node is more than the parent (or middle) node. By setting the current node to be equal to the right node you are iterating through the tree in order from least to largest value. Note that if cur.right() doesn't exist then cur will be set to null and therefore not execute the while loop.

Answer 2

I submitted my code, it was successful.
https://leetcode.com/problems/binary-search-tree-iterator/
You can find it here.
https://github.com/yan-khonski-it/bst/blob/master/bst-core/src/main/java/com/yk/training/bst/iterators/BSTIterator.java
Explanation.
You have to use inorder which first visits the left sub-tree, then visit the current node, and then the right sub-tree, so you will iterate through all the elements in the ascending order.
Now, you have stack, which holds all the node that you should return in next call in the correct order.
next will remove the last element from the stack. Now you check the current node, if it has right subtree. If so, you need to iterate though left elements of the right subtree.
/**
* Find next node to be returned in {#link #next()}.
* Push it to stack.
*/
public void navigateLeftSubtree() {
stack.push(currentNode);
while (currentNode.left != null) {
currentNode = currentNode.left;
stack.push(currentNode);
}
}
In this case, (right sub tree is present for current node), you should put the right child of the current node into the stack. You don't want to put the current into the stack, if you have already visited it.
public int next() {
currentNode = stack.pop();
final int currentValue = currentNode.value;
if (currentNode.right != null) {
// Push root of the right subtree into stack.
currentNode = currentNode.right;
navigateLeftSubtree();
}
return currentValue;
}