共计 1319 个字符，预计需要花费 4 分钟才能阅读完成。
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]
]
难度：medium
题目：给定二叉树，返回其之字型层次遍历结点值。
思路：层次遍历
Runtime: 1 ms, faster than 90.32% of Java online submissions for Binary Tree Zigzag Level Order Traversal.Memory Usage: 37.1 MB, less than 100.00% of Java online submissions for Binary Tree Zigzag Level Order Traversal.
/**
* 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>> result = new ArrayList<>();
if (null == root) {
return result;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
int currentLevelCount = 1, nextLevelCount = 0;
boolean direction = true;
while (!queue.isEmpty()) {
List<Integer> curLevelElems = new ArrayList<>(currentLevelCount);
for (int i = 0; i < currentLevelCount; i++) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.add(node.left);
nextLevelCount++;
}
if (node.right != null) {
queue.add(node.right);
nextLevelCount++;
}
if (direction) {
curLevelElems.add(node.val);
} else {
curLevelElems.add(0, node.val);
}
}
result.add(curLevelElems);
currentLevelCount = nextLevelCount;
nextLevelCount = 0;
direction = !direction;
}
return result;
}
}
