LeetCode | Binary Tree Level Order Traversal
2506 ワード
タイトル:
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.
考え方:
タイトルは二叉木の幅優先遍歴に似ており,1つのキューで実現できる.しかし、キュー内の各ノードには、数値表示階層(level)を追加する必要があります.隣接する階層を区別するだけなので、階層をlevel%2またはbool値に簡略化して表すことができます.
コード: /**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
TreeNode * q1[10000];
bool q2[10000];
int begin=0;
int end=0;
if(root == NULL)
{
vector<vector<int> > v;
return v;
}
else
{
vector<vector<int> > v;
q1[end] = root;
q2[end++] = true;
bool level = true;
vector<int> * tmp = new vector<int>();
while(begin!=end)
{
TreeNode * p = q1[begin];
bool cur = q2[begin++];
if(cur != level)
{
v.push_back(*tmp);
delete tmp;
tmp = new vector<int>();
level = !level;
}
if(cur == level)
{
if(p->left != NULL)
{
q1[end] = p->left;
q2[end++] = !cur;
}
if(p->right != NULL)
{
q1[end] = p->right;
q2[end++] = !cur;
}
tmp->push_back(p->val);
}
}
v.push_back(*tmp);
return v;
}
}
};
3
/ \
9 20
/ \
15 7
[
[3],
[9,20],
[15,7]
]
タイトルは二叉木の幅優先遍歴に似ており,1つのキューで実現できる.しかし、キュー内の各ノードには、数値表示階層(level)を追加する必要があります.隣接する階層を区別するだけなので、階層をlevel%2またはbool値に簡略化して表すことができます.
コード: /**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
TreeNode * q1[10000];
bool q2[10000];
int begin=0;
int end=0;
if(root == NULL)
{
vector<vector<int> > v;
return v;
}
else
{
vector<vector<int> > v;
q1[end] = root;
q2[end++] = true;
bool level = true;
vector<int> * tmp = new vector<int>();
while(begin!=end)
{
TreeNode * p = q1[begin];
bool cur = q2[begin++];
if(cur != level)
{
v.push_back(*tmp);
delete tmp;
tmp = new vector<int>();
level = !level;
}
if(cur == level)
{
if(p->left != NULL)
{
q1[end] = p->left;
q2[end++] = !cur;
}
if(p->right != NULL)
{
q1[end] = p->right;
q2[end++] = !cur;
}
tmp->push_back(p->val);
}
}
v.push_back(*tmp);
return v;
}
}
};
/**
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
TreeNode * q1[10000];
bool q2[10000];
int begin=0;
int end=0;
if(root == NULL)
{
vector<vector<int> > v;
return v;
}
else
{
vector<vector<int> > v;
q1[end] = root;
q2[end++] = true;
bool level = true;
vector<int> * tmp = new vector<int>();
while(begin!=end)
{
TreeNode * p = q1[begin];
bool cur = q2[begin++];
if(cur != level)
{
v.push_back(*tmp);
delete tmp;
tmp = new vector<int>();
level = !level;
}
if(cur == level)
{
if(p->left != NULL)
{
q1[end] = p->left;
q2[end++] = !cur;
}
if(p->right != NULL)
{
q1[end] = p->right;
q2[end++] = !cur;
}
tmp->push_back(p->val);
}
}
v.push_back(*tmp);
return v;
}
}
};