Level-order Traversal

https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/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).

Thoughts

要求按层返回结果，自然想到按层遍历。搁一行逆转结果，那bfs时记录下是否该逆转然后对需要逆转的用stack存储结果即可。

Code

/**
 * 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;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        boolean clock = true;
        while (!queue.isEmpty()) {
            int length = queue.size();
            Stack<TreeNode> stack = new Stack<>();
            List<Integer> subres = new ArrayList<>();
            for (int i = 0; i < length; i++) {
                TreeNode node = queue.poll();
                if (clock) {
                    subres.add(node.val);
                } else {
                    stack.push(node);
                }
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
            if (!clock) {
                while (!stack.isEmpty()) {
                    subres.add(stack.pop().val);
                }
                clock = true;
            } else {
                clock = false;
            }
            res.add(subres);
        }
        return res;
    }
}

Analysis

时间复杂度为O(n)