Leetcode -- parity tree

-------------------------------------------- Pure brush notes , Not your own way of solving problems -------------------------------------
、 Title Description ：
If a binary tree satisfies the following conditions , It can be called Even tree ：

 The subscript of the layer where the root node of the binary tree is located is  0 , The layer of the child node of the root is subscript  1 , The layer of the root's grandson is subscript  2 , And so on .
 Even subscript   The values of all nodes on the layer are   p.   Integers , From left to right   Strictly increasing 
 Odd subscript   The values of all nodes on the layer are   accidentally   Integers , From left to right   Strictly decreasing 

Give you the root node of the binary tree , If the binary tree is Even tree , Then return to true , Otherwise return to false .

answer ：

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    bool isEvenOddTree(TreeNode* root) {
    
        //  Breadth first traversal   This topic breadth first search is indeed more appropriate   It can match the current level 
        queue<TreeNode*> qu;
        qu.push(root);
        int level = 0;
        while (!qu.empty()) {
    
            int size = qu.size();
            // cout<<size<<endl;
            int prev = level % 2 == 0 ? INT_MIN : INT_MAX;
            for (int i = 0; i < size; i++) {
    
                // Returns the first element 
                TreeNode * node = qu.front();
                // Pop up the first element 
                qu.pop();
                cout<< i <<endl;
                int value = node->val;
                if (level % 2 == value % 2) {
    
                    return false;
                }
                if ((level % 2 == 0 && value <= prev) || (level % 2 == 1 && value >= prev)) {
    
                    return false;
                }
                prev = value;
                if (node->left != nullptr) {
    
                //  Current level breadth traversal 
                    qu.push(node->left);
                }
                if (node->right != nullptr) {
    
                    qu.push(node->right);
                }
            }
            level++;
        }
        return true;
    }
};