5.1.5 Binary Tree Level Order Traversal II

文章目录

• ​​1.题目​​
• ​​2.代码​​

1.题目

• ​​题目要求​​
• 思路：​​将5.1.5二叉树层序遍历​​进行reverse翻转即可
• eg：
  Example 1:
  
  Input: root = [3,9,20,null,null,15,7]
  Output: [[15,7],[9,20],[3]]

2.代码

递归版
class Solution{
public:
  vector<vector<int>> levelOrderBottom(TreeNode* root){
    vector<vector<int>> result;
    int level=0;
    tranversfunction();
    reverse(result.begin(),result.end());//多一行
    return result;
  }
  void tranversfunction(vector<vector<int>>& result, int level, TreeNode* root)
  {
    if (!root)  return;
    result[level].push_back(root->val);
    if (result.size() == level) result.push_back(vector<int>());
    if (root->left) tranversfunction(result,level+1,root->left);
    if (root->right) tranversfunction(result,level+1,root->right);
  }
};

class Solution{
public:
  vector<vector<int>> levelOrderBottom(TreeNode* root){
    vector<vector<int>> result;
    deque<TreeNode*> deq;
    dep.push(root);
    while (!dep.empty())
    {
      vector<int> tempvec;
      for (int i=dep.size();i>0;--i)
      {
        TreeNode* node=dep.front();
        dep.pop();
        tempvec.push_back(node->val);
        if (node->left) dep.push(node->left);
        if (node->right) dep.push(node->right);
      }
      result.push_back(tempvec);
    }
    
    reverse(result.begin(), result.end());//多一行
    return result;

  }


};