• Binary Tree Level Order Traversal


    Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

    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:每一次一层中的所有结点插入到一个vector<TreeNode*>中,然后将这一层中的vector<TreeNode*>插入到包含所有层的vector<vector<TreeNode*> >中。从根结点开始,每次取出一层,访问该层的所以结点,如果其左右子树不为空,则将其左右子树加入到vector<TreeNode*>中,最后在将这一层插入到vector<vector<TreeNode*>>中,每次都是针对最后一层进行的操作。

    C++代码实现:

    #include<iostream>
    #include<new>
    #include<vector>
    using namespace std;
    
    //Definition for binary tree
    struct TreeNode
    {
        int val;
        TreeNode *left;
        TreeNode *right;
        TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    };
    
    class Solution
    {
    public:
        vector<vector<int> > levelOrder(TreeNode *root)
        {
            vector<vector<int> > vec;
            vector<vector<TreeNode*> > rvec;
            size_t i=0;
            if(root==NULL)
                return vector<vector<int> >();
            vec.push_back({root->val});
            rvec.push_back({root});
            vector<int> v1;
            vector<TreeNode*> v2;
            while(vec.size()&&rvec.size())
            {
                v1.clear();
                v2.clear();
                for(i=0; i<rvec[rvec.size()-1].size(); i++)
                {
                    cout<<"level : "<<vec.size()<<"val: "<<vec[rvec.size()-1][i]<<endl;
                    TreeNode *tmp=rvec[rvec.size()-1][i];
                    cout<<"level : "<<vec.size()<<"node: "<<tmp->val<<endl;
                    if(tmp->left)
                    {
                        v1.push_back(tmp->left->val);
                        v2.push_back(tmp->left);
                    }
                    if(tmp->right)
                    {
                        v1.push_back(tmp->right->val);
                        v2.push_back(tmp->right);
                    }
                }
                if(!v1.empty()&&!v2.empty())
                {
                    vec.push_back(v1);
                    rvec.push_back(v2);
                }
                else
                    break;
            }
            return vec;
        }
        void createTree(TreeNode *&root)
        {
            int i;
            cin>>i;
            if(i!=0)
            {
                root=new TreeNode(i);
                if(root==NULL)
                    return;
                createTree(root->left);
                createTree(root->right);
            }
        }
    };
    
    int main()
    {
        Solution s;
        TreeNode *root;
        s.createTree(root);
        vector<vector<int> > vec=s.levelOrder(root);
        for(auto a:vec)
        {
            for(auto v:a)
                cout<<v<<" ";
            cout<<endl;
        }
    }

    运行结果:

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  • 原文地址:https://www.cnblogs.com/wuchanming/p/4099537.html
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