# Tree Coloring - Amazon Top Interview Questions

### Problem Statement :

You are given a binary tree root where the value of each node represents its color. In the tree there are at most 2 colors. Return whether it's possible to swap the colors of the nodes any number of times so that no two adjacent nodes have the same color.

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [1, null, null], [1, [0, null, [0, null, null]], [0, null, null]]]

Output

True

Explanation

We can color the root with 0, the next level 1s, third level 0s and fourth level 1s

Example 2

Input

root = [5, null, [9, [5, null, null], [9, null, null]]]

Output

False

There's no way to color the nodes so that no two adjacent nodes have the same color.

### Solution :

Solution in C++ :

/**
* class Tree {
*     public:
*         int val;
*         Tree *left;
*         Tree *right;
* };
*/

int dfs(Tree *root, int color) {
if (!root) return 0;
return color + dfs(root->left, 1 - color) + dfs(root->right, 1 - color);
}

void countNodes(Tree *root, int firstColor, int &firstCount, int &secondCount) {
if (!root) return;

if (root->val == firstColor)
++firstCount;
else
++secondCount;

countNodes(root->left, firstColor, firstCount, secondCount);
countNodes(root->right, firstColor, firstCount, secondCount);
}

bool solve(Tree *root) {
if (!root) return true;

int req = dfs(root, 1), firstColor = root->val, firstCount = 0, secondCount = 0;
countNodes(root, firstColor, firstCount, secondCount);
return req == firstCount || req == secondCount;
}

Solution in Python :

# class Tree:
#     def __init__(self, val, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def solve(self, root):
if not root:
return True
evens = 0  # how many nodes are in even layers
tot = 0  # how many total nodes there are
firstCnt = 0  # how many nodes have the "first" color
first = root.val  # we assign the root node's color to be the first color
q = collections.deque([(root, 0)])
while q:
curr, lvl = q.popleft()
tot += 1
if curr.val == first:
firstCnt += 1
if lvl % 2:
evens += 1
if curr.left:
q.append((curr.left, lvl + 1))
if curr.right:
q.append((curr.right, lvl + 1))
return evens in [
firstCnt,
tot - firstCnt,
]  # the even layers can get the first or second color

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