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
/ \
9 20
…/ \
…15 7
返回锯齿形层序遍历如下:
[
[3],
[20,9],
[15,7]
]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal
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思路:bfs题,处理下是左往右还是右往左就行了
/**
* 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) {
List<List<Integer>> res = new LinkedList<List<Integer>>();
if (root == null) {
return res;}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
//offer和put的区别,队列满offer返回false,put抛出异常
queue.offer(root);
boolean isOrderLeft = true;
while (!queue.isEmpty()) {
Deque<Integer> levelList = new LinkedList<Integer>();
int size = queue.size();
for (int i = 0; i < size; ++i) {
TreeNode curNode = queue.poll();
if (isOrderLeft) {
levelList.offerLast(curNode.val);
} else {
levelList.offerFirst(curNode.val);
}
if (curNode.left != null) {
queue.offer(curNode.left);
}
if (curNode.right != null) {
queue.offer(curNode.right);
}
}
res.add(new LinkedList<Integer>(levelList));
isOrderLeft = !isOrderLeft;
}
return res;
}
}