Delete Leaves With a Given Value - Leetcode 1325 - Python

FIRST!

SECOND!!

THIRD

Fourth

fifth

Sixty Nine!

Thank you so much for the daily problems.

Is there any reason why would we use bfs in this case?

Thank you!

Ayo the thumbnail😂

Thank You 🍃

420 everyday

420 everyday

Dunno why is it medium. Pretty straightforward solution.

Accepted Java Iterative Solution using map and stack:

class Solution {
public TreeNode removeLeafNodes(TreeNode root, int target) {

Map<TreeNode, TreeNode> parentMap = new HashMap();
Stack<TreeNode> stk = new Stack();
stk.push(root);

while (!stk.isEmpty()) {

TreeNode node = stk.pop();

if (node.right == null && node.left == null && node.val == target) { // delete this node

TreeNode parent = parentMap.get(node);

if (parent != null && parent.right == node) { // update parent right pointer to null

parentMap.remove(node);
parent.right = null;

}

if (parent != null && parent.left == node) { // update parent left pointer to null

parentMap.remove(node);
parent.left = null;

}

} else {

boolean added = false;

if (node.right != null && !parentMap.containsKey(node.right)) {

added = true;
stk.push(node); // add back current node to stack
stk.push(node.right);
parentMap.put(node.right, node);
}

if (node.left != null && !parentMap.containsKey(node.left)) {

if (!added) {

stk.push(node); // if not added ealier then add back
}

stk.push(node.left);
parentMap.put(node.left, node);
}

}

}

//if root only is pending and having value == target
if (root.left == null && root.right == null && root.val == target) {

return null;
}

return root;
}
}

honestly I doubt we will ever have chance to implement tree traversal with iterative approach...

thanks for the explanation

Yes, The leaf node check should be done post order (not pre order) -- so that it will reach till parent