[Leetcode] Binary tree level order traversal二叉树层次遍历
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]
]
confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
/**
* 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)
{
vector<vector<int>> res;
vector<int> levelNode;
queue<TreeNode *> Q;
if(root) Q.push(root);
int count=; //下一层元素的个数
int levCount=; //当前层元素个数,初始为第一层
while( !Q.empty())
{
TreeNode *cur=Q.front();
levelNode.push_back(cur->val);
Q.pop();
levCount--;
if(cur->left)
{
Q.push(cur->left);
count++;
}
if(cur->right)
{
Q.push(cur->right);
count++;
}
if(levCount==)
{
res.push_back(levelNode);
levCount=count;
count=;
levelNode.clear(); //清空levelNode,为下层
}
}
return res;
}
};
方法二:
思路:遍历完一层以后,队列中节点的个数就是二叉树下一层的节点数。实时更新队列中节点的个数,每层的遍历。
/**
* 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)
{
vector<vector<int>> res;
queue<TreeNode *> Q;
if(root) Q.push(root); while( !Q.empty())
{
int count=;
int levCount=Q.size();
vector<int> levNode; //遍历当前层
while(count<levCount)
{
TreeNode *curNode=Q.front();
Q.pop();
levNode.push_back(curNode->val);
if(curNode->left)
Q.push(curNode->left);
if(curNode->right)
Q.push(curNode->right);
count++;
}
res.push_back(levNode);
}
return res;
}
};
/**
* 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)
{
vector<vector<int>> res;
queue<TreeNode *> Q;
if(!root) return res;
Q.push(root);
Q.push(NULL);
vector<int> levNode; //存放每层的结点的值 while( !Q.empty())
{
TreeNode *cur=Q.front();
Q.pop();
if(cur)
{
levNode.push_back(cur->val);
if(cur->left)
Q.push(cur->left);
if(cur->right)
Q.push(cur->right);
}
else
{
res.push_back(levNode);
levNode.clear();
if( !Q.empty())
Q.push(NULL);
}
}
return res;
}
};
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