华为企业网站建设分析seo关键词排名优化方法

103. 二叉树的锯齿形层序遍历

给你二叉树的根节点 root ，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）。

示例 1：输入：root = [3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]示例 2：输入：root = [1]
输出：[[1]]示例 3：输入：root = []
输出：[]提示：树中节点数目在范围 [0, 2000] 内
-100 <= Node.val <= 100

题解：

方法一：按层模拟BFS

/*** 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 void reverse(List<Integer> list){int size = list.size();int tmp[] = new int[size];for(int i=0;i<size;i++){tmp[i] = list.get(i);}int index = 0;for(int i=size-1;i>=0;i--){list.set(index,tmp[i]);index++;}}public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<>();if(root == null){return res;}Queue<TreeNode> queue = new LinkedList<>();boolean flag = true; // true代表->   false代表<-List<Integer> first = new ArrayList<>();first.add(root.val);if(root.left != null)queue.offer(root.left);if(root.right != null)queue.offer(root.right);res.add(first);while(!queue.isEmpty()){List<Integer> tmp = new ArrayList<>();int count = queue.size();while(count > 0){TreeNode node = queue.poll();if(node.left != null)queue.offer(node.left);if(node.right != null)queue.offer(node.right);tmp.add(node.val);count--;}flag = !flag;if(!flag){//对此时取到的tmp顺序取反reverse(tmp);}res.add(tmp);}return res;}
}

方法二：双端队列+奇偶

/*** 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) {List<List<Integer>> res = new ArrayList<>();if(root == null){return res;}Queue<TreeNode> queue = new LinkedList<>();int len = 1;// 奇数代表->   偶数代表<-List<Integer> first = new LinkedList<>();first.add(root.val);if(root.left != null)queue.offer(root.left);if(root.right != null)queue.offer(root.right);res.add(first);len++;while(!queue.isEmpty()){// 队列依旧是传统队列，但是每一个加入到res中的小list都是用双端形式，从而形式上实现双端队列List<Integer> tmp = new LinkedList<>();// 也是因为链表形式相较于数组形式更利于反转int count = queue.size();while(count > 0){TreeNode node = queue.poll();if(node.left != null)queue.add(node.left);      if(node.right != null)queue.offer(node.right);if(len % 2 == 0){tmp.addFirst(node.val); }else{tmp.addLast(node.val);}count--;}res.add(tmp);len++;}return res;}
}