5.1.6 二叉树的之字形层序遍历

文章目录

• ​​1.题目​​
• ​​2.代码​​

1.题目

• ​​题目要求​​
• 思路：最简单直接的方法就是利用层序遍历，并使用一个变量 cnt 来统计当前的层数（从0开始），将所有的奇数层的结点值进行翻转一下即可
• eg

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
 

return its zigzag level order traversal as:

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

2.代码

class Solution{
  vector<vector<int>> zigzagLevelOrder(TreeNode* root)
  {
    vector<vector<int>> result;
    int level=0;
    transefunction(result,level,root);
    return result;
  }
  
  void transefunction(vector<vector<int>>& result, int level, TreeNode* node)
  {
    if (!node) return;
    if (result.size() == level) result.push_back(vector<int>());
    int singleordouble=0;
    if (singleordouble%2 == 0)
      tempresult[level].push_back(node->val);
    else
      tempresult[level].insert(tempresult[level].begin(), node->val);
    if (node->left) transefunction(result,level+1,node->left);
    if (node->right) transefunction(result,level+1,node->right);
  }

};


class Solution{
  vector<vector<int>> zigzagLevelOrder(TreeNode* root)
  {
    vector<vector<int>> result;
    deque<TreeNode*> deq;
    int sigleordouble=0;
    deq.push(root);
    while(!deq.empty())
    {
      vector<int> tmpresult;
      for (int i=deq.size();i>0;--i)
      {
        TreeNode* pre=deq.front();
        deq.pop();
        tmpresult.push_back(pre->val);
        if (pre->left) deq.push(pre->left);
        if (pre->right)  deq.push(pre->right);
      }
      if (sigleordouble%2!=0)
        reverse(tmpresult.begin().tmpresult.end());
      result.push_back(tmpresult);
      sigleordouble++;
    }
    
    return result;
  }
};