标签：leetcode bfs 层序遍历 stack queue
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.
thinking：
和普通的二叉树层序遍历原理一样，借助queue实现，先得到从上到下的层序遍历结果，再借助stack翻转一下结果即可
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);
}
};leetcode || 107、Binary Tree Level Order Traversal II
标签：leetcode bfs 层序遍历 stack queue
原文地址：http://blog.csdn.net/hustyangju/article/details/45217187
