Balance Tree Coins with the Fewest Adjacent Transfers

### Problem A binary tree has n nodes and exactly n coins in total; each node stores its current coin count in val. One move transfers one coin across one parent-child edge in either direction. Return the minimum moves needed to leave exactly one coin at every node. ```cpp class Solution { public: int distributeCoins(TreeNode* root); }; ``` ### Examples **Example 1:** ```text Input: root = [1,0,2] Output: 2 Explanation: Move one coin from the right child to the root, then from the root to the left child. ``` **Example 2:** ```text Input: root = [0,0,null,3] Output: 3 Explanation: The bottom left grandchild sends two coins upward to its parent, and that parent sends one to the root. ``` ### Constraints - The tree contains n nodes, with `1 <= n <= 100`. - `0 <= Node.val <= n`. - All node values sum to n. ### Contract considerations Node values represent coin counts, not costs or immutable labels. The reference computes flows without modifying those values. ```hint Read the boundary condition For one subtree, how many coins must cross its single connection to the parent? ``` ### Boundary reminders Node values represent coin counts, not costs or immutable labels. The reference computes flows without modifying those values.