Group Binary Tree Leaves by Removal Round

# Group Binary Tree Leaves by Removal Round Given the root of a binary tree, repeatedly collect all current leaves, then remove those leaves. Continue until the tree is empty. Return the node values grouped by removal round, with the first round first. A leaf is a node with no children. All leaves in one round are collected before the next round begins. Values within the same round may appear in any order. ### Function `vector<vector<int>> findLeaves(TreeNode* root)` Use the supplied `TreeNode` objects and their `val`, `left`, and `right` fields. Do not replace the tree interface with an array-input function. ### Example 1 ```text Input tree: [8,4,9,null,6] Output: [[6,9],[4],[8]] ``` Initially, nodes `6` and `9` are leaves. Removing them makes node `4` a leaf. The root `8` is removed last. `[[9,6],[4],[8]]` is also valid. ### Example 2 ```text Input tree: [7] Output: [[7]] ``` ### Constraints - The tree has between `1` and `100` nodes. - `-100 <= Node.val <= 100`. ### Clarifications The level-order arrays describe trees for the examples; the function receives a root pointer. Different nodes may have equal values, so every node contributes one occurrence to its round. The rounds themselves must remain in chronological order. ### What a Strong Answer Covers - The meaning of simultaneous removal within a round. - Correct groups for uneven branches, a single node, and repeated values. - Why every node appears in its correct round exactly once. - Time and auxiliary-space bounds.