请实现一个函数按照之字形打印二叉树,即第一行按照从左到右的顺序打印,第二层按照从右至左的顺序打印,第三行按照从左到右的顺序打印,其他行以此类推。
package Tree;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
/**
* 请实现一个函数按照之字形打印二叉树,即第一行按照从左到右的顺序打印,第二层按照从右至左的顺序打印,第三行按照从左到右的顺序打印,其他行以此类推。
* 思路:
* 先按层次输出二叉树
* 判断奇数层和偶数层
* 反转arrayList
*/
public class Solution9 {
public static void main(String[] args) {
int[] array = {1, 2, 3, 4, 5, 6, 7, 8, 9};
Solution9 Solution9 = new Solution9();
TreeNode treeNode = Solution9.createBinaryTreeByArray(array, 0);
for (ArrayList list :
Solution9.Print(treeNode)) {
System.out.println(list);
}
}
/**
* 之字形打印二叉树
* 用reserve反转,时间复杂度高
*
* @param pRoot
* @return
*/
public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
//arrayLists存储结果
ArrayList<ArrayList<Integer>> arrayLists = new ArrayList<>();
if (pRoot == null) {
return arrayLists;
}
ArrayList<Integer> arrayList = new ArrayList<>();
//使用队列,先进先出
Queue<TreeNode> queue = new LinkedList<>();
queue.add(pRoot);
int start = 0;
int end = 1;
boolean leftToRight = true;
while (!queue.isEmpty()) {
TreeNode temp = queue.remove();
//添加到本行的arrayList
arrayList.add(temp.val);
start++;
//每打印一个节点,就把此节点的下一层的左右节点加入队列,并记录下一层要打印的个数
if (temp.left != null) {
queue.add(temp.left);
}
if (temp.right != null) {
queue.add(temp.right);
}
if (start == end) {
start = 0;
end = queue.size();
if (leftToRight) {
arrayLists.add(arrayList);
} else {
arrayLists.add(reverse(arrayList));
}
leftToRight = !leftToRight;
arrayList = new ArrayList<>();
}
}
return arrayLists;
}
/**
* 反转
*
* @param arrayList
* @return
*/
private ArrayList<Integer> reverse(ArrayList<Integer> arrayList) {
ArrayList<Integer> arrayList1 = new ArrayList<>();
for (int i = arrayList.size() - 1; i >= 0; i--) {
arrayList1.add(arrayList.remove(i));
}
return arrayList1;
}
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
/**
* 数据转二叉树
*
* @param array
* @param index
* @return
*/
private TreeNode createBinaryTreeByArray(int[] array, int index) {
TreeNode tn = null;
if (index < array.length) {
int value = array[index];
tn = new TreeNode(value);
tn.left = createBinaryTreeByArray(array, 2 * index + 1);
tn.right = createBinaryTreeByArray(array, 2 * index + 2);
return tn;
}
return tn;
}
}