Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 29

Description

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 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.

Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 1 <= Node.val <= 106

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  bool isEvenOddTree(TreeNode* root) {
    queue<TreeNode*> q;
    q.push(root);
    int level = 0;
    while(q.size()) {
      int sz = q.size();
      int prev = (level % 2) ? INT_MAX : INT_MIN;
      auto op = (level % 2) ? [](int a, int b){ return a <= b ;} : [](int a, int b){ return a >= b; };
      for(int _ = 0; _ < sz; ++_) {
        auto current = q.front();
        q.pop();
        if(current->val % 2 == level % 2) return false;
        if(op(prev, current->val))return false;
        prev = current->val;
        if(current->left) q.push(current->left);
        if(current->right) q.push(current->right);
      }
      level += 1;
    }
    return true;
  }
};

// Accepted
// 106/106 cases passed (125 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 76.25 % of cpp submissions (149.2 MB)