In Distribute Coins in Binary Tree (LC 979), a move takes one coin across one edge. Every node must end with exactly one coin. The total number of coins equals the number of nodes.

Look at one subtree and the edge that joins it to its parent. Coins can enter or leave the subtree only through that edge. If the subtree holds 3 more coins than nodes, 3 coins must leave through that edge. If it holds 2 fewer, 2 coins must enter. So the subtree needs exactly one number to describe what crosses the edge: its coins minus its nodes. This is excess(node). A positive excess is spare coins. A negative excess is a debt.

Take root = [0, 3, 0]. The left child holds 3 coins and is one node, so it returns 3 - 1 = +2. The right child holds 0 coins, so it returns 0 - 1 = -1. At the root, moves grows by |+2| + |-1| = 3. The root returns 0 + 2 + (-1) - 1 = 0, because the coins equal the nodes. The answer is 3: two coins leave the left child, one stays at the root and one goes on to the right child.

There is a lower bound for each edge. If the subtree below an edge has excess e, then at least |e| coins must cross that edge. A move crosses only one edge, so the fewest moves is at least the sum of |e| over all edges.

The bound can be reached. Settle the tree in postorder. When a node is handled, its children have already passed their surplus up or asked for coins, so the node can pass spare coins to a child that has a debt, or send them up. No coin has to cross an edge in both directions, and no edge is crossed more than |e| times. So the sum is not only a lower bound but also the answer.

This is why the code adds abs(left) + abs(right) to moves at each node. The two terms are the two child edges. Each edge is counted once, at its upper node.

A second example: root = [1, 0, 0, null, 3]. The node 3 returns +2. Its parent, the left child 0, adds |0| + |2| = 2 to moves and returns 0 + 0 + 2 - 1 = +1. The right child 0 returns -1. The root adds |+1| + |-1| = 2, so moves is 4. The root returns 0.

The first trap is abs in the return statement. The function must return the signed excess. A subtree that is short of coins must return a negative number. If the function returns abs(node.val + left + right - 1), a debt of 1 comes back as a surplus of 1, and the parent counts coins that do not exist. For root = [2, 0, 2, 0] the correct answer is 4, and the code with abs in the return gives 2. Take abs only when adding to moves.

The second trap is the - 1. Every node keeps one coin, so the excess of a subtree is its coins minus its nodes, and the node itself counts as one. A leaf with one coin must return 0. Without the - 1, that leaf returns 1, as if it had a spare coin. For root = [3, 0, 0] the correct answer is 2, and the code without - 1 gives 0.

A third trap is to return moves from the function. The function returns the excess, and moves is a separate counter kept outside, so it needs nonlocal. The answer is moves after the call on the root ends.

Each node is visited once, and each visit does constant work. The time is O(N). The space is O(H) for the recursion stack, where H is the height.

This card is built on Bottom-Up DFS. Each call returns one number for its subtree, and the parent combines the numbers of its children. The difference is that the number is signed, and the parent uses it twice: once to add a cost to moves for the edge below, and once to build its own number.

Compare with the card Tree DP (Max Path Sum / Diameter). There you look for one best path that may bend at any node, and the answer is stored in global_max. Here you add a cost for every edge, and the answer is the total in moves. Every edge is counted, not one chosen path.

Other problems on this card have the same shape: a subtree reports one number, and the parent pays a cost for the edge from that number. Minimum Fuel Cost to Report to the Capital (LC 2477) is one of them.