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)。
xml
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
java
/**
* 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
java
/**
* 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;
}
}