102. Binary Tree Level Order Traversal

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102. 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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

题意：逐层打印二叉树上的节点

代码一、采用二叉树的层次遍历的思想进行

vector<vector<int>> levelOrder(TreeNode* root) {        //采用的是二叉树的层次遍历思想，其中每一层的数据用一个临时temp的vector<int>来保存        if(NULL == root)            return vector<vector<int>>();                queue<TreeNode*> que;        que.push(root);        int countLenNodes = 1; //记录的是每一层上的节点数        vector<vector<int>> result;        vector<int> temp;        while(!que.empty())        {            TreeNode* tempNode = que.front();            temp.push_back(tempNode->val);            que.pop();            countLenNodes--;                        if(tempNode->left != NULL)                que.push(tempNode->left);            if(tempNode->right != NULL)                que.push(tempNode->right);            if(countLenNodes == 0)            {                result.push_back(temp);                temp.resize(0); //或者temp.clear()均可以                countLenNodes = que.size();            }        }        return result;    }

方法二、前序递归遍历构建vector

vector<vector<int>> result;        void buildVector(TreeNode* root, int depth)    {        if(NULL == root)            return;        if(result.size() == depth) //？？？当开创新的层次的时候，采用该语句进行申请空间            result.push_back(vector<int>());                result[depth].push_back(root->val);        buildVector(root->left,depth + 1);        buildVector(root->right,depth + 1);    }        vector<vector<int>> levelOrder(TreeNode* root) {        buildVector(root,0);        return result;    }

方法三、传统的两个队列的解决方案

class Solution {  public:      vector<vector<int>> levelOrder(TreeNode* root)       {                        vector<vector<int>> result;      queue<TreeNode*> q;        if (root == NULL)      {          return result;      }      q.push(root);      vector<int> le_temp;        while(!q.empty())      {            le_temp.clear();          queue<TreeNode*> level;                        int size = q.size();              for(int i = 0; i < size; ++i)          {              TreeNode* temp = q.front();              q.pop();              if(temp->left)              {                  level.push(temp->left);              }              if(temp->right)              {                  level.push(temp->right);                                }              le_temp.push_back(temp->val);          }            while(!level.empty())          {                q.push(level.front());                 level.pop();          }          result.push_back(le_temp);      }        return result;                }  };

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