LeetCode:Binary Tree Level Order Traversal I II

LeetCode: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]
]                                                                      本文地址

队列辅助层序遍历，队列中插入NULL作为层与层之间的间隔，注意处理队列里最后的NULL时，不能再把它入队列以免形成死循环

/**
* 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) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<vector<int> >res;
if(root == NULL)return res;
queue<TreeNode*> myqueue;
myqueue.push(root);
myqueue.push(NULL);//NULL是层与层之间间隔标志
vector<int> level;
while(myqueue.empty() == false)
{
TreeNode *p = myqueue.front();
myqueue.pop();
if(p != NULL)
{
level.push_back(p->val);
if(p->left)myqueue.push(p->left);
if(p->right)myqueue.push(p->right);
}
else
{
res.push_back(level);
if(myqueue.empty() == false)
{
level.clear();
myqueue.push(NULL);
}
}
}
return res;
}
};

LeetCode:Binary Tree Level Order Traversal II

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],
]

和上一题差不多，只是需要把最后遍历结果数组翻转一下

/**
* 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> > levelOrderBottom(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<vector<int> >res;
if(root == NULL)return res;
queue<TreeNode*> myqueue;
myqueue.push(root);
myqueue.push(NULL);//NULL是层与层之间间隔标志
vector<int> level;
while(myqueue.empty() == false)
{
TreeNode *p = myqueue.front();
myqueue.pop();
if(p != NULL)
{
level.push_back(p->val);
if(p->left)myqueue.push(p->left);
if(p->right)myqueue.push(p->right);
}
else
{
res.push_back(level);
if(myqueue.empty() == false)
{
level.clear();
myqueue.push(NULL);
}
}
}
reverse(res.begin(), res.end());
return res;
}
};

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