20. Even Odd Tree

A binary tree is named Even-Odd if it meets the following conditions:

• The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2, etc.

• For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).

• For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right).

Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.

Example 1:

Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
Output: true
Explanation: The node values on each level are:
Level 0: [1]
Level 1: [10,4]
Level 2: [3,7,9]
Level 3: [12,8,6,2]
Since levels 0 and 2 are all odd and increasing, and levels 1 and 3 are all even and decreasing, the tree is Even-Odd.

Example 2:

Input: root = [5,4,2,3,3,7]
Output: false
Explanation: The node values on each level are:
Level 0: [5]
Level 1: [4,2]
Level 2: [3,3,7]
Node values in the level 2 must be in strictly increasing order, so the tree is not Even-Odd.

Example 3:

Input: root = [5,9,1,3,5,7]
Output: false
Explanation: Node values in the level 1 should be even integers.

Example 4:

Input: root = [1]
Output: true

Example 5:

Input: root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
Output: true

Solution: (BFS and Array)

class Solution
{
public:
    bool isEvenOddTree(TreeNode *root)
    {

        queue<TreeNode *> q;
        q.push(root);

        int level = 0;

        while (!q.empty())
        {

            int s = q.size();
            vector<int> v;
            while (s--)
            {
                TreeNode *t = q.front();
                q.pop();
                v.push_back(t->val);
                if (t->left)
                {
                    q.push(t->left);
                }

                if (t->right)
                {
                    q.push(t->right);
                }
            }

            int prev = (level%2==0)? -1 : INT_MAX; 
            for (int i = 0; i < v.size(); i++)
            {
                if (level % 2 == 0)
                {
                    if (v[i] % 2 == 0 || v[i] <= prev)
                    {
                        return false;
                    }
                }

                if (level % 2 != 0)
                {
                    if (v[i] % 2 != 0 || v[i]>=prev)
                    {
                        return false;
                    }
                }

                prev = v[i];
            }
            level++;
        }

        return true;
    }
};

Solution: (Optimized BFS)

class Solution
{
public:
    bool isEvenOddTree(TreeNode *root)
    {

        queue<TreeNode *> q;
        q.push(root);

        int level = 0;
        
        while (!q.empty())
        {

            int s = q.size();
            int prev = (level % 2 == 0) ? -1 : INT_MAX;
            
            while (s--)
            {
                TreeNode *t = q.front();
                q.pop();

                if (t->left)
                {
                    q.push(t->left);
                }

                if (t->right)
                {
                    q.push(t->right);
                }

                if (level % 2 == 0)
                {
                    if (t->val % 2 == 0 || t->val <= prev)
                    {
                        return false;
                    }
                }

                if (level % 2 != 0)
                {
                    if (t->val % 2 != 0 || t->val >= prev)
                    {
                        return false;
                    }
                }

                prev = t->val;
            }

            level++;
        }

        return true;
    }
};