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题目及测试
package pid103;
import java.util.List;
/* 二叉树的锯齿形层次遍历
给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
例如:
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回锯齿形层次遍历如下:
[
[3],
[20,9],
[15,7]
]
}*/
public class main {
public static void main(String[] args) {
Object[] x=new Object[]{1,2,3,4,null,null,5};
BinaryTree tree=new BinaryTree(x);
tree.printTree(tree.root);
test(tree.root);
Object[] x2=new Object[]{3,9,20,null,null,15,7};
BinaryTree tree2=new BinaryTree(x2);
tree2.printTree(tree2.root);
test(tree2.root);
}
private static void test(TreeNode ito) {
Solution solution = new Solution();
List<List<Integer>> rtn;
long begin = System.currentTimeMillis();
rtn = solution.zigzagLevelOrder(ito);//执行程序
long end = System.currentTimeMillis();
System.out.println("rtn=" );
for (List<Integer> list : rtn) {
for (Integer integer : list) {
System.out.print(integer+" ");
}
System.out.println();
}
System.out.println();
System.out.println("耗时:" + (end - begin) + "ms");
System.out.println("-------------------");
}
}
解法1(成功,2ms,较快)
使用典型的二叉树层次化遍历的方法,设置一个队列逐个放入层次化的二叉树,设置num,next变量控制层数,然后设置isLTR变量,表明该层的数据从左往右还是从右往左,如果从右往左,则每层遍历的节点由于是从左往右,则塞入list的顺序是插入头部,导致最右的数据在list最左边
package pid103;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
Queue<TreeNode> queue=new LinkedList<>();
List<List<Integer>> result=new ArrayList<>();
LinkedList<Integer> list=new LinkedList<>();//一排的integer
if(root==null){
return result;
}
queue.add(root);
int num=1;
int next=0;
boolean isLTR=true;//是否这一排顺序从左到右
while(!queue.isEmpty()){
TreeNode now=queue.poll();
if(now.left!=null){
queue.add(now.left);
next++;
}
if(now.right!=null){
queue.add(now.right);
next++;
}
if(isLTR){
list.add(now.val);
}
else{
list.addFirst(now.val);
}
num--;
if(num==0){
num=next;
next=0;
isLTR=!isLTR;
result.add(list);
list=new LinkedList<>();
}
}
return result;
}
}