Binary Tree Zigzag Level Order Traversal

发布于 2025-02-22 13:01:29 字数 2902 浏览 0 评论 0 收藏 0

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leetcode: Binary Tree Zigzag Level Order Traversal | LeetCode OJ
lintcode: (71) 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).

Example
Given binary tree {3,9,20,#,#,15,7},

  3
   / \
  9  20
  /  \
   15   7


return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

题解 1 - 队列

二叉树的广度优先遍历使用队列非常容易实现，这道题要求的是蛇形遍历，我们可以发现奇数行的遍历仍然可以按照广度优先遍历的方式实现，而对于偶数行，只要翻转一下就好了。

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */

public class Solution {
  /**
   * @param root: The root of binary tree.
   * @return: A list of lists of integer include
   *      the zigzag level order traversal of its nodes' values
   */
  public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
    if (root == null) return result;

    boolean odd = true;
    Queue<TreeNode> q = new LinkedList<TreeNode>();
    q.offer(root);
    while (!q.isEmpty()) {
      // level traversal
      int qLen = q.size();
      ArrayList<Integer> level = new ArrayList<Integer>();
      for (int i = 0; i < qLen; i++) {
        TreeNode node = q.poll();
        level.add(node.val);
        if (node.left != null) q.offer(node.left);
        if (node.right != null) q.offer(node.right);
      }
      // add level order reverse for even
      if (odd) {
        result.add(level);
      } else {
        Collections.reverse(level);
        result.add(level);
      }
      // flip odd and even
      odd = !odd;
    }

    return result;
  }
}