Binary Tree Level Order Traversal II

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#csdn #二叉树 #层次遍历 #迭代 #算法

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本文详细介绍了使用递归和迭代两种方法解决二叉树层次遍历问题，包括算法思想和具体实现步骤。通过实例分析，帮助读者理解不同方法的优缺点，并掌握二叉树层次遍历的基本技巧。

题目：

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).

For example:

return its bottom-up level order traversal as:

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

算法思想：

1，首先想到的是递归，对于一个结点，将左子树和右子树返回的结果合并起来，再加上根结点这一层。

2，第二种思路是迭代，将每一层的结点分别存储在二维数组的一行，然后逆向输出每一行。

vector > levelOrderBottom(TreeNode *root) 
{
    if (root == NULL) return vector>();
    
    vector> res_left = levelOrderBottom(root->left);
    vector> res_right = levelOrderBottom(root->right);

   
    if (res_right.size() < res_left.size())
    {
        int k = res_left.size()-res_right.size();
        
        for (int i = 0; i < res_right.size(); i++)
        {
            for (int j = 0; j < res_right[i].size(); j++)
                res_left[k+i].push_back(res_right[i][j]);
        }
        
        res_left.push_back(vector(1, root->val));
        
        return res_left;
    }
    else
    {
        int k = res_right.size()-res_left.size();
        
        for (int i = 0; i < res_left.size(); i++)
        {
            for (int j = 0; j < res_left[i].size(); j++)
                res_right[k+i].insert(res_right[k+i].begin()+j, res_left[i][j]);
        }
       
        res_right.push_back(vector(1, root->val));
        
        return res_right;
    }
}

vector > levelOrderBottom(TreeNode *root) 
{
    if (root == NULL) return vector>();
    
    vector*> record;
    vector> res;
    vector* p;
    vector* p1;
    
    p = new vector;
    p->push_back(root);
    record.push_back(p);
    
    while (true)
    {
        p1 = new vector;
        
        for(int i = 0; i < p->size(); i++)
        {
            if ((*p)[i]->left != NULL)
                p1->push_back((*p)[i]->left);
            if ((*p)[i]->right != NULL)
                p1->push_back((*p)[i]->right);
        }
        
        if (p1->size() == 0) break;
        
        record.push_back(p1);
        p = p1;
    }
    
    for(int i = record.size()-1; i >= 0; i--)
    {
        vector temp;
        
        for(int j = 0; j < record[i]->size(); j++)
            temp.push_back((*record[i])[j]->val);
        
        res.push_back(temp);
    }
    
    return res;
}