题解 | #牛群的二叉树排序#
牛群的二叉树排序
https://www.nowcoder.com/practice/a3a8756cbb13493ab4cf5d73c853d5cd
题目考察的知识点
考察二叉树的深度优先搜索
题目解答方法的文字分析
需要统计0 1 的个数,随后利用队列从中取值,分别构建左右子树,将对应的值填入即可。
本题解析所用的编程语言
使用Java代码解答
完整且正确的编程代码
import java.util.*;
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* public TreeNode(int val) {
* this.val = val;
* }
* }
*/
public class Solution {
/**
* 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整型一维数组
* @return TreeNode类
*/
public TreeNode sortCowsTree (int[] cows) {
// write code here
int zeros = 0;
int ones = 0;
for (int i : cows) {
if (i == 0) {
zeros++;
} else {
ones++;
}
}
TreeNode root = new TreeNode(-1);
if (zeros > 0) {
TreeNode zero = new TreeNode(0);
root.left = zero;
sortCows(zero, zeros - 1, 0);
}
if (ones > 0) {
TreeNode one = new TreeNode(1);
root.right = one;
sortCows(one, ones - 1, 1);
}
return root;
}
private void sortCows(TreeNode root, int num, int flag) {
Queue<TreeNode> queue = new ArrayDeque<>();
queue.offer(root);
while (num > 0) {
TreeNode temp = queue.poll();
TreeNode lTemp = new TreeNode(flag);
temp.left = lTemp;
queue.offer(lTemp);
num--;
if (num > 0) {
TreeNode rTemp = new TreeNode(flag);
temp.right = rTemp;
queue.offer(rTemp);
num--;
}
}
}
}

