LeetCode之“树”：Binary Tree Level Order Traversal && Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal

　　题目链接

　　题目要求：

　　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]
]

　　这道题利用宽度优先搜索就可以了，具体程序（8ms）如下：

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> retVec;
        if(!root)
            return retVec;

        retVec.push_back(vector<int>{root->val});
        queue<TreeNode *> que;
        que.push(root);
        while(true)
        {
            vector<int> vec;
            queue<TreeNode *> q;
            while(!que.empty())
            {
                TreeNode *tree = que.front();
                que.pop();
                if(tree->left)
                {
                    vec.push_back((tree->left)->val);
                    q.push(tree->left);
                }
                if(tree->right)
                {
                    vec.push_back((tree->right)->val);
                    q.push(tree->right);
                }
            }
            
            if(!q.empty())
            {
                que = q;
                retVec.push_back(vec);
            }
            else
                break;
        }
        
        return retVec;
    }
};

Binary Tree Level Order Traversal II

　　题目链接

　　题目要求：

　　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:
　　Given binary tree {3,9,20,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

　　return its bottom-up level order traversal as:

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

　　这道题基本更上边的题目一样，我们只需要将上题的结果retVec最后再反转一下就可以了，即添加如下一行即可：

reverse(retVec.begin(), retVec.end());

　　这样的程序只要8ms，但下边基本一样的程序却要64ms：

/**
 * 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<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> retVec;
        if(!root)
            return retVec;

        retVec.insert(retVec.begin(), vector<int>{root->val});
        queue<TreeNode *> que;
        que.push(root);
        while(true)
        {
            vector<int> vec;
            queue<TreeNode *> q;
            while(!que.empty())
            {
                TreeNode *tree = que.front();
                que.pop();
                if(tree->left)
                {
                    vec.push_back((tree->left)->val);
                    q.push(tree->left);
                }
                if(tree->right)
                {
                    vec.push_back((tree->right)->val);
                    q.push(tree->right);
                }
            }
            
            if(!q.empty())
            {
                que = q;
                retVec.insert(retVec.begin(), vec);
            }
            else
                break;
        }
            
        return retVec;
    }
};