[LeetCode] 103. Binary Tree Zigzag Level Order Traversal

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]
]

二叉树的锯齿形层序遍历。

题意是给一个二叉树，对其进行层序遍历，但是在遍历每一层的时候要一次从左开始一次从右开始。

思路和102题的做法 - 层序遍历没两样，唯一需要多一个变量记录到底是从左遍历还是从右遍历。

时间O(n)

空间O(n) - 记录output

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function (root) {
    var res = [];
    // corner case
    if (root === null) {
        return res;
    }

    // normal case
    var queue = [];
    queue.push(root);
    var x = true;
    while (queue.length > 0) {
        var size = queue.length;
        var list = [];
        for (var i = 0; i < size; i++) {
            var cur = queue.shift();
            if (x) {
                list.push(cur.val);
            } else {
                list.unshift(cur.val);
            }
            if (cur.left !== null) {
                queue.push(cur.left);
            }
            if (cur.right !== null) {
                queue.push(cur.right);
            }
        }
        res.push(list);
        x = !x;
    }
    return res;
};

Java实现

/**
 * 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<>();
        // corner case
        if (root == null) {
            return res;
        }

        // normal case
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        boolean x = true;
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (x) {
                    list.add(cur.val);
                } else {
                    list.add(0, cur.val);
                }
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            res.add(list);
            x = !x;
        }
        return res;
    }
}

LeetCode 题目总结