[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] ]
简单的层次遍历,两个栈
/**
* 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
vector<vector<int> > ret;
ret.clear();
if(root == NULL)
return ret;
stack<TreeNode*> S;
stack<TreeNode*> T;
S.push(root);
while(!S.empty()){
vector<int> tmp;
while(!S.empty()){
TreeNode* p = S.top();
S.pop();
tmp.push_back(p->val);
if(p->left)
T.push(p->left);
if(p->right)
T.push(p->right);
}
ret.push_back(tmp);
if(T.empty())
return ret;
while(!T.empty())
{
TreeNode *p = T.top();
T.pop();
S.push(p);
}
}
}
};
又写了一个队列的,用NULL指针把层次分隔开。
/**
* 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
vector<vector<int> > ret;
ret.clear();
if(root == NULL)
return ret;
queue<TreeNode*> S;
S.push(root);
S.push(NULL);
vector<int> tmp;
while(!S.empty()){
//travesal current level
TreeNode* p = S.front();
S.pop();
if(p!=NULL)
{
tmp.push_back(p->val);
if(p->left)
S.push(p->left);
if(p->right)
S.push(p->right);
}else{
if(!tmp.empty())
{
S.push(NULL);
ret.push_back(tmp);
tmp.clear();
}
}
}
return ret;
}
};