0103. 二叉树的锯齿形层序遍历【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - BFS + 方向交替

1. 📝 题目描述

• leetcode

给你二叉树的根节点 root，返回其节点值的 锯齿形层序遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

---

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]

示例 2：

输入：root = [1]
输出：[[1]]

示例 3：

输入：root = []
输出：[]

---

提示：

• 树中节点数目在范围 [0, 2000] 内
• -100 <= Node.val <= 100

2. 🎯 s.1 - BFS + 方向交替

c

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** zigzagLevelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if (!root) { *returnColumnSizes = NULL; return NULL; }

    int** ans = (int**)malloc(2000 * sizeof(int*));
    *returnColumnSizes = (int*)malloc(2000 * sizeof(int));
    struct TreeNode* queue[2001];
    int front = 0, rear = 0;
    queue[rear++] = root;
    int leftToRight = 1;

    while (front < rear) {
        int size = rear - front;
        ans[*returnSize] = (int*)malloc(size * sizeof(int));
        (*returnColumnSizes)[*returnSize] = size;
        for (int i = 0; i < size; i++) {
            struct TreeNode* node = queue[front++];
            int idx = leftToRight ? i : size - 1 - i;
            ans[*returnSize][idx] = node->val;
            if (node->left) queue[rear++] = node->left;
            if (node->right) queue[rear++] = node->right;
        }
        leftToRight = !leftToRight;
        (*returnSize)++;
    }

    return ans;
}

js

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function (root) {
  if (!root) return []
  const ans = []
  const queue = [root]
  let leftToRight = true

  while (queue.length) {
    const size = queue.length
    const level = new Array(size)
    for (let i = 0; i < size; i++) {
      const node = queue.shift()
      // 根据方向决定插入位置
      level[leftToRight ? i : size - 1 - i] = node.val
      if (node.left) queue.push(node.left)
      if (node.right) queue.push(node.right)
    }
    ans.push(level)
    leftToRight = !leftToRight
  }

  return ans
}

py

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        ans = []
        queue = deque([root])
        left_to_right = True

        while queue:
            size = len(queue)
            level = [0] * size
            for i in range(size):
                node = queue.popleft()
                idx = i if left_to_right else size - 1 - i
                level[idx] = node.val
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            ans.append(level)
            left_to_right = not left_to_right

        return ans

• 时间复杂度：$O(n)$，其中 $n$ 是二叉树的节点数，每个节点入队出队各一次
• 空间复杂度：$O(n)$，队列最多存储一层节点

算法思路：

• 标准 BFS 层序遍历，额外维护一个布尔值 leftToRight 标记当前层的填充方向
• 每层预分配定长数组 level，根据方向决定节点值的插入位置：正序时放 i，逆序时放 size - 1 - i
• 每层结束后翻转方向标记