binary-tree-level-order-traversal I、II——输出二叉树的数字序列
I
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;
if(root==NULL) return res;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
int n=q.size();
vector<int> v;
for(int i=;i<n;i++){
TreeNode *cur=q.front();
q.pop();
v.push_back(cur->val);
if(cur->left!=NULL)
q.push(cur->left);
if(cur->right!=NULL)
q.push(cur->right);
}
res.push_back(v);
}
return res;
}
};
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],
]
confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
先将结果v存入stack中，最后在从stack倒入res形成倒序，未找到其他好的方法
/**
* 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) {
vector<vector<int>> res;
if(root==NULL) return res;
stack<vector<int>> s;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
int n=q.size();
vector<int> v;
for(int i=;i<n;i++){
TreeNode *cur=q.front();
q.pop();
v.push_back(cur->val);
if(cur->left!=NULL)
q.push(cur->left);
if(cur->right!=NULL)
q.push(cur->right); }
s.push(v);
}
while(!s.empty()){
res.push_back(s.top());
s.pop();
}
return res;
}
};
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