0107. 二叉树的层序遍历 II【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - BFS + 结果翻转

1. 📝 题目描述

• leetcode

给你二叉树的根节点 root，返回其节点值 自底向上的层序遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

---

示例 1：

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

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
• -1000 <= Node.val <= 1000

2. 🎯 s.1 - BFS + 结果翻转

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** levelOrderBottom(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if (!root) { *returnColumnSizes = NULL; return NULL; }

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

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

    // 翻转结果数组
    for (int i = 0, j = *returnSize - 1; i < j; i++, j--) {
        int* tmpRow = ans[i]; ans[i] = ans[j]; ans[j] = tmpRow;
        int tmpCol = colSizes[i]; colSizes[i] = colSizes[j]; colSizes[j] = tmpCol;
    }

    *returnColumnSizes = colSizes;
    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 levelOrderBottom = function (root) {
  if (!root) return []
  const ans = []
  const queue = [root]

  while (queue.length) {
    const size = queue.length
    const level = []
    for (let i = 0; i < size; i++) {
      const node = queue.shift()
      level.push(node.val)
      if (node.left) queue.push(node.left)
      if (node.right) queue.push(node.right)
    }
    ans.push(level)
  }

  return ans.reverse()
}

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 levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        ans = []
        queue = deque([root])

        while queue:
            level = []
            for _ in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            ans.append(level)

        return ans[::-1]

• 时间复杂度：$O(n)$，其中 $n$ 是二叉树的节点数
• 空间复杂度：$O(n)$，队列最多存储一层节点

算法思路：

• 与标准层序遍历（0102）完全一致，先自顶向下 BFS 收集每层结果
• 最后将结果数组整体翻转，即可得到自底向上的层序遍历