5.1.5二叉树层序遍历

文章目录

• ​​1.题目​​
• ​​2.代码​​

1.题目

• ​​题目要求​​

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7
 

return its level order traversal as:

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

• 思路：
  ​​迭代法​​：使用队列，更好理解
  ​递归法​​

2.代码

迭代法
class Solution 
{
public:
  vector<vector<int>> levelOrder(TreeNode* root)
  {
    vector<vector<int>> result;
    //层序遍历多了个level
    int level=0;
    transfunction(result,level,root);
    return result;
  }
  void transfunction(vector<vector<int>>& tmpvec, int tmpllevel, TreeNode* node)
  {
    if (!node) return;
    if (tmpvec.size() == tmpllevel) return tmpllevel.push_back(vector<int>);
    tmpvec[tmpllevel].push_back(node->val);
    if (node->left) transfunction(tmpvec,tmpllevel+1,node->left);
    if (node->right) transfunction(tmpvec,tmpllevel+1,node->right);
  }
}; 


递归版
class Solution{
public:
  vector<vector<int>> levelOrder(TreeNode* root){
  vector<vector<int>> result;
  deque<TreeNode* > deq;
  deq.push(root);
  while (!dep.empty())
  {
    vector<int> subresult;
    for (int i=deq.size();i>0;--i)
    {
      TreeNode* deqnode=deq.front();
      deq.pop();
      subresult.push_back(deqnode);
      if (deqnode->left) dep.push(deqnode->left);
      if (deqnode->right) dep.push(deqnode->right);
    }
    result.push_back(subresult);
    
  }
  return result;
}


};