[LeetCode] 107. Binary Tree Leve

原题

层序遍历，从自底向上按层输出。 左→右→中
解法一 :
DFS，求出自顶向下的，最后返回时反转一下。

class Solution
{
public:
  vector<vector<int>> res;
  vector<vector<int>> levelOrderBottom(TreeNode *root)
  {
    int level = 0;
    dfs(root, level);
    reverse(res.begin(), res.end());
    return res;
  }
  void dfs(TreeNode *root, int level)
  {
    if (root == NULL)
      return;
    if (level == res.size())
      res.push_back({});
    dfs(root->left, level + 1);
    dfs(root->right, level + 1);
    res[level].push_back(root->val);
  }
};

解法二：
BFS

class Solution
{
public:
  vector<vector<int>> res;
  vector<vector<int>> levelOrderBottom(TreeNode *root)
  {
    if (root == NULL)
      return res;
    int level = 0;
    queue<TreeNode *> que;
    que.push(root);
    while (!que.empty())
    {
      vector<int> cur;
      int level = que.size();
      for (int i = 0; i < level; i++)
      {
        TreeNode *temp = que.front();
        que.pop();
        cur.push_back(temp->val);

        if (temp->left != NULL)
          que.push(temp->left);
        if (temp->right != NULL)
          que.push(temp->right);
      }
      res.insert(res.begin(), cur);
    }
    return res;
  }
};