题解 | #牛群的二叉树排序#

题目考察的知识点：二叉树的遍历

题目解答方法的文字分析：构造左子树，构造右子树；用队列进行构造，然后合并左右子树。

本题解析所用的编程语言：c++

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param cows int整型vector 
     * @return TreeNode类
     */
    TreeNode* leftnode(int count, int x)
    {
        queue<TreeNode*> que;
        TreeNode* root= new TreeNode(x);
        TreeNode* t;
        que.push(root);
        while (--count > 0)
        {
            t = new TreeNode(x);
            que.push(t);
            que.front()->left = t;
            --count;
            if (count > 0)
            {
                t = new TreeNode(x);
                que.push(t);
                que.front()->right = t;
            }
            que.pop();
        }
        return root;
    }
    TreeNode* sortCowsTree(vector<int>& cows) {
        // write code here
        int left = 0;
        int right = 0;
        for (auto& x : cows)
        {
            if (x == 0)
                ++left;
            else
                ++right;
        }
        TreeNode* root = new TreeNode(-1);
        if (left > 0) root->left = leftnode(left, 0);
        if (right > 0) root->right = leftnode(right, 1);
        return root;
    }
};