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;
}
};

View Code