(This question has been seen in the interviews of the following companies: Google)

#### Tree Game

**Data Structure:**

```
class TreeNode {
TreeNode parent; //parent node
TreeNode left; //left child
TreeNode right; //right child
}
```

**Question:**

Two people in a game, player scores by claiming nodes in binary tree, tree node class as shown above.

The player who eventually owns more nodes wins the game.

Player A and B each claims a node at first.

After the first round, a player will only be able to claim a node adjacent to any node owned by himself.

A tree node is adjacent to its parent, left right and right child.

A node owned cannot be re-claimed.

End game when all nodes are owned.

If player A gets the first claim at node N, find whether it is possible for player B to win.

If yes, find out which node player B should claim at his first move.

**Follow up**:

if player B takes the first hand instead, which node should he pick?

```
class TreeNode {
TreeNode parent; //parent node
TreeNode left; //left child
TreeNode right; //right child
}
```

**Solution: **

```
class TreeNode {
int id;
TreeNode left;
TreeNode right;
TreeNode parent;
TreeNode(int id) {
this.id = id;
}
}
/*
The opponent's first move on Node N divides the tree into 3 components - left subtree, right subtree and parent branch of N.
Your best move is to claim a node adjacent to Node N at the biggest component.
Function countNodes() counts the sizes of 3 components. Function win() finds the largest component, whose size is your score.
*/
public static boolean win(TreeNode root, TreeNode n) { //N is the first move by opponent
int sizeParent = countNodes(n.parent, n); //size of parent component
int sizeLeft = countNodes(n.left, n); //size of left subtree component
int sizeRight = countNodes(n.right, n); //size of right subtree component
int myScore = Math.max(Math.max(sizeParent, sizeLeft), sizeRight); //I take the biggest component
int treeSize = 1 + sizeParent + sizeLeft + sizeRight;
int opponentScore = treeSize - myScore; //opponent takes remaining nodes
System.out.print("my best score is " + myScore + "/" + treeSize + ". ");
if(myScore > opponentScore) {
TreeNode bestmove = myScore == sizeParent ? n.parent: myScore == sizeLeft ? n.left : n.right;
System.out.println("my first move on " + bestmove.id);
}
return myScore > opponentScore;
}
private static int countNodes(TreeNode node, TreeNode ignore) {
if(node == null) return 0;
int count = 1;
if(node.parent != ignore) {
count += countNodes(node.parent, node);
}
if(node.left != ignore) {
count += countNodes(node.left, node);
}
if(node.right != ignore) {
count += countNodes(node.right, node);
}
return count;
}
```

**Solution to the followup: **

```
/*
To find the best move we must count the sizes of 3 adjacent components for every node in the tree.
If exists a node whose biggest adjacent component is smaller than treesize/2, the node is your best move, cuz your opponent won't be able to get a score higher than treesize/2.
In some cases there is no winning play, like on a tree with 2 nodes.
We can store the component sizes of every node in a 2-dimensional cache.
1st dimension: node
2nd dimension: which component (parent, left or right)
value: size of component
*/
public static int bestMove(TreeNode root) {
if(root == null) return -1;
if(root.left == null && root.right == null) return root.id;
// map stores size of every component
// each node at most has 3 components - to its left, to its right, to its top (parent)
// Map
```>
Map> components = new HashMap<>();
TreeNode dummy = new TreeNode(-1);
dummy.left = root;
//calculate size of child components for all nodes
getComponentSize(dummy, root, components);
int treeSize = components.get(dummy).get(root);
for(TreeNode node: components.keySet()) {
int maxComponentSize = 0; //maximum score possible for opponent
for(int size: components.get(node).values()) {
if(size > maxComponentSize) maxComponentSize = size;
}
if(treeSize / 2.0 > maxComponentSize) { //opponent cannot get half of the tree. You win.
return node.id; //best first move
}
}
return -1; //no winning play
}
private static int getComponentSize(TreeNode n, TreeNode branch, Map> components) {
if(n == null || branch == null) return 0;
if(components.containsKey(n) && components.get(n).containsKey(branch)) {
return components.get(n).get(branch); // component size of a branch from node n (n excluded)
}
// a node n has 3 branches at most - parent, left, right
if(!components.containsKey(n)) {
components.put(n, new HashMap<>());
}
int size = 1; // 1 is the size of TreeNode branch
if(branch == n.left || branch == n.right) {
//size of the subtree 'branch' is size(branch.left) + size(branch.right) + 1
size += getComponentSize(branch, branch.left, components);
size += getComponentSize(branch, branch.right, components);
} else { //else when (branch == n.parent)
//see the tree from left-side or right-side view (see parent as a child; see one of the children as parent)
//size of the component is size(branch.parent) + size(branch.left/right child)
size += getComponentSize(branch, branch.parent, components);
size += branch.left == n ? getComponentSize(branch, branch.right, components) : getComponentSize(branch, branch.left, components);
}
components.get(n).put(branch, size); // cache the result of recursion
getComponentSize(n, n.parent, components); // calculate size of parent component for current node
return size;
}

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