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[Leetcode]107. Binary Tree Level Order Traversal II(C++)

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题目描述

题目链接：107. Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

例子

Input: [3,9,20,null,null,15,7]
Output:

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

解题思路

通过 BFS 按层输出节点，最后逆转得到的向量即可，代码如下： 

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
#include <algorithm>
#include <queue>
#include <vector>

class Solution {
public:
  std::vector<std::vector<int>> levelOrderBottom(TreeNode *root) {
    std::vector<std::vector<int>> result;
    if (root == nullptr)
      return result;

    std::queue<TreeNode *> q;
    q.psuh(root);
    while (!q.empty()) {
      std::queue<TreeNode *> next_q;
      std::vector<int> level;
      while (!q.empty()) {
        TreeNode *node = q.front();
        q.pop();

        level.push_back(node->val);
        if (node->left)
          next_q.push(node->left);
        if (node->right)
          next_q.push(node->right);
      }
      result.push_back(level);
      level.clear();
      std::swap(q, next_q);
    }

    std::reverse(result.begin(), result.end());

    return result;
  }
};
  • 时间复杂度: O(n)
  • 空间复杂度: O(w) — 每层最大宽度

GitHub 代码同步地址： 107.BinaryTreeLevelOrderTraversalIi.cpp

其他题目：
GitHub: Leetcode-C++-Solution
博客： Leetcode-Solutions

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