【LeetCode】 103. Binary Tree Zigzag Level Order Traversal 二叉树的锯齿形层次遍历(Medium)(JAVA)
题目地址: https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/
题目描述:
Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
题目大意
给定一个二叉树,返回其节点值的锯齿形层次遍历。(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)。
解题方法
和上一题解法相同:【LeetCode】 102. Binary Tree Level Order Traversal 二叉树的层序遍历(Medium)(JAVA)
只需要每遍历一层,反转一次即可
/**
* 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 ArrayList<>();
if (root == null) return res;
List<Integer> cur = new ArrayList<>();
List<TreeNode> nodes = new ArrayList<>();
nodes.add(root);
int count = 1;
boolean flag = true;
while (nodes.size() > 0) {
TreeNode node = nodes.remove(0);
count--;
if (node.left != null) nodes.add(node.left);
if (node.right != null) nodes.add(node.right);
if (flag) {
cur.add(node.val);
} else {
cur.add(0, node.val);
}
if (count == 0) {
count = nodes.size();
res.add(new ArrayList<>(cur));
cur = new ArrayList<>();
flag = !flag;
}
}
return res;
}
}
执行用时 : 2 ms, 在所有 Java 提交中击败了 27.12% 的用户
内存消耗 : 38.3 MB, 在所有 Java 提交中击败了 7.41% 的用户