https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/
给定一个二叉树,返回其节点值的锯齿形层序遍历。(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<br /> / \<br /> 9 20<br /> / \<br /> 15 7<br />返回锯齿形层序遍历如下:
[
[3],
[20,9],
[15,7]
]
层序遍历,对当前层数判断一下是奇数还是偶数就行,偶数正序,奇数翻转list再存入就可以
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
if(root == null){
return new ArrayList<>();
}
//定义一个队列用来操作节点
Queue<TreeNode> queue = new LinkedList<>();
List<List<Integer>> res = new ArrayList<>();
queue.offer(root);
//用来记录当前节点个数
int curnum = 1;
//记录层数
int flag = 0;
while(!queue.isEmpty()){
ArrayList<Integer> list = new ArrayList<>();
int nextsum = 0;
//用来记录下一层节点个数
for(int i = 0; i<curnum; i++){
TreeNode node = queue.poll();
list.add(node.val);
if(node.left!=null){
queue.offer(node.left);
nextsum++;
}
if(node.right!=null){
queue.offer(node.right);
nextsum++;
}
}
if(flag%2 == 0){
res.add(list);
}else{
Collections.reverse(list);
res.add(list);
}
flag++;
curnum = nextsum;
}
return res;
}
}