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)。
给定二叉树 [3,9,20,null,null,15,7],
返回锯齿形层序遍历如下:
[
[3],
[20,9],
[15,7]
]
链接:https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal/
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(root == null) { //二叉树为空,直接返回空结果
return res;
}
ArrayDeque<TreeNode> deque = new ArrayDeque<TreeNode>();
deque.add(root);
leverOrder(res, deque, 1);
return res;
}
/*
* 参数列表说明:
* res 结果集
* deque 双向队列存储当前层次遍历的结点
* flag 记录当前遍历的层数,用来控制层序遍历的方向
*/
void leverOrder(List<List<Integer>> res,ArrayDeque<TreeNode> deque,int flag) {
int size = deque.size();
if(size == 0) {
return;
}
ArrayList<Integer> arrayList = new ArrayList<Integer>();
if(flag == 1) {
while(size-- != 0) {
TreeNode root = deque.pollFirst();
arrayList.add(root.val);
if(root.left != null) {
deque.offer(root.left);
}
if(root.right != null) {
deque.offer(root.right);
}
}
//加入结果集,递归下一层
res.add(arrayList);
leverOrder(res, deque, 2);
}else {
while(size-- != 0) {
TreeNode root = deque.pollLast();
arrayList.add(root.val);
if(root.right != null) {
deque.offerFirst(root.right);
}
if(root.left != null) {
deque.offerFirst(root.left);
}
}
//加入结果集,//递归下一层
res.add(arrayList);
leverOrder(res, deque, 1);
}
}
}
原文:https://www.cnblogs.com/fyy-f/p/14172136.html