• [LeetCode] Binary Tree Right Side View


    Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

    For example:
    Given the following binary tree,

       1            <---
     /   
    2     3         <---
          
      5     4       <---
    

    You should return [1, 3, 4].

    Credits:

    分析:层次遍历,取最右面的元素:

    方法一:双q法:

    class Solution {
        public:
            vector<int>  rightSideView(TreeNode *root)
            {
                queue<TreeNode*> q1;
                queue<TreeNode*> q2;
                vector<int>  res;
    
                if(root != NULL)
                {
                    q1.push(root);
                }
    
                while(!q1.empty())
                {
                    TreeNode * p = q1.front();
                    q1.pop();
    
                    if(p->left)
                        q2.push(p->left);
                    if(p->right)
                        q2.push(p->right);
    
                    if(q1.empty() /*&& !q2.empty()*/)
                    {
                        res.push_back(p->val);
                        swap(q1, q2);
                    }
                }
                return res;
            }
    };

    方法二:单q+ NULL标记level结束法

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
        public:
            vector<int> rightSideView(TreeNode* root) {
    
                vector<int> rtn;
                if(root == NULL)
                    return rtn;
    
                queue<TreeNode*> q;
    
                TreeNode* pre = NULL;
    
                q.push(root);
                q.push(NULL);//mark the end of this level
    
                while(!q.empty())
                {   
                    TreeNode * p = q.front();
                    q.pop();
    
                    if(p == NULL) 
                    {   
                        rtn.push_back(pre->val);
    
                        if(!q.empty())// some elements still exist in q.
                            q.push(NULL);//mark end of this level
    
                    }   
                    else
                    {   
                        if(p->left)
                            q.push(p->left);
                        if(p->right)
                            q.push(p->right);
                    }   
                    pre = p;
                }
    
                return rtn;
            }
    };
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  • 原文地址:https://www.cnblogs.com/diegodu/p/4582981.html
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