leetcode---binary-tree-level-order-traversal-ii---树层次遍历

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.

OJ’s Binary Tree Serialization:

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> > levelOrderBottom(TreeNode *root)
{
queue<TreeNode *> q;
vector<vector<int> > ans;
if(!root)
return ans;
q.push(root);
stack<vector<int> > s;

while(!q.empty())
{
vector<int> v;
queue<TreeNode *> q1;
while(!q.empty())
{
TreeNode *p = q.front();
q.pop();
v.push_back(p->val);
if(p->left)
q1.push(p->left);
if(p->right)
q1.push(p->right);
}
q = q1;
if(!v.empty())
s.push(v);
}

while(!s.empty())
{
vector<int> v = s.top();
s.pop();
ans.push_back(v);
}

return ans;
}
};