Description
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]
,
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
Solution
Level-order-traversal, time O(n), space O(n)
稍微变种一下就ok啦。
/**
* 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>> levels = new LinkedList<>();
if (root == null) {
return levels;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
boolean inReverse = false;
while (!queue.isEmpty()) {
int levelSize = queue.size();
List<Integer> level = new LinkedList<>();
while (levelSize-- > 0) {
TreeNode node = queue.poll();
if (inReverse) {
level.add(0, node.val); // insert to the begining
} else {
level.add(node.val);
}
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
levels.add(level);
inReverse = !inReverse;
}
return levels;
}
}
Two-stack, time O(n), space O(n)
用两个stack来回折腾。
/**
* 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>> levels = new LinkedList<>();
if (root == null) {
return levels;
}
Stack<TreeNode> order = new Stack<>();
Stack<TreeNode> reverse = new Stack<>();
order.push(root);
List<Integer> level = new LinkedList<>();
while (!order.empty()) {
while (!order.empty()) {
TreeNode node = order.pop();
level.add(node.val);
if (node.left != null) reverse.push(node.left);
if (node.right != null) reverse.push(node.right);
}
levels.add(new LinkedList<>(level));
level.clear();
if (reverse.empty()) {
break;
}
while (!reverse.empty()) {
TreeNode node = reverse.pop();
level.add(node.val);
if (node.right != null) order.push(node.right);
if (node.left != null) order.push(node.left);
}
levels.add(new LinkedList<>(level));
level.clear();
}
return levels;
}
}