「力扣」第 103 题:二叉树的锯齿形层次遍历(中等)
传送门:103. 二叉树的锯齿形层次遍历。
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)。
例如:
给定二叉树[3,9,20,null,null,15,7]
,3 / \ 9 20 / \ 15 7
返回锯齿形层次遍历如下:
[ [3], [20,9], [15,7] ]
Java 代码:
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
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<>();
if (root == null) {
return res;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
boolean direction = true;
while (!queue.isEmpty()) {
// 当前这一层遍历的节点集合
List<Integer> curList = new ArrayList<>();
// 特别注意:每一次只能处理上一轮入队列的的元素,
// 所以要将上一轮入队列的元素个数先存一下
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode curNode = queue.poll();
if (direction) {
curList.add(curNode.val);
} else {
curList.add(0, curNode.val);
}
// 处理每一个元素都一样,都要考虑一下左右子树
if (curNode.left != null) {
queue.add(curNode.left);
}
if (curNode.right != null) {
queue.add(curNode.right);
}
}
// 改换方向
direction = !direction;
res.add(curList);
}
return res;
}
public static void main(String[] args) {
TreeNode node1 = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node3 = new TreeNode(3);
TreeNode node4 = new TreeNode(4);
TreeNode node5 = new TreeNode(5);
TreeNode node6 = new TreeNode(6);
TreeNode node7 = new TreeNode(7);
node1.left = node2;
node1.right = node3;
node2.left = node4;
node2.right = node5;
node3.left = node6;
node3.right = node7;
Solution solution = new Solution();
List<List<Integer>> zigzagLevelOrder = solution.zigzagLevelOrder(node1);
zigzagLevelOrder.forEach(System.out::println);
}
}
Python 代码: