problem:
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.
Hide Tags Tree Breadth-first Search 題意:層序周遊二叉樹,從最後一層往上開始按層輸出
thinking:
和普通的二叉樹層序周遊原理一樣,借助queue實作,先得到從上到下的層序周遊結果,再借助stack翻轉一下結果即可
code:
class Solution {
private:
vector<vector<int> > ret;
stack<vector<int> > _stack;
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) {
ret.clear();
if(root==NULL)
return ret;
queue<TreeNode *> tmp_queue;
tmp_queue.push(root);
level_order(tmp_queue);
while(!_stack.empty())
{
vector<int> tmp=_stack.top();
ret.push_back(tmp);
_stack.pop();
}
return ret;
}
void level_order(queue<TreeNode *> queue1)
{
if(queue1.empty())
return;
vector<int> array;
queue<TreeNode *> queue2;
while(!queue1.empty())
{
TreeNode *tmp=queue1.front();
array.push_back(tmp->val);
queue1.pop();
if(tmp->left!=NULL)
queue2.push(tmp->left);
if(tmp->right!=NULL)
queue2.push(tmp->right);
}
_stack.push(array);
level_order(queue2);
}
};