【LeetCode】Binary Tree Level Order Traversal II

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}"

.
思路：参考Binary Tree Level Order Traversal。但是要注意最后要翻转。

reverse(res.begin(),res.end());

代码参考下面。当然还可以使用栈来实现。这样顺序就要换一下。

class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int> > res;
vector<int> onelev;
if(root==NULL)return res;
queue<TreeNode *> que;
que.push(root);
que.push(NULL);//第一层
TreeNode *tr;
while(1){
tr=que.front();
que.pop();

if(tr!=NULL){
onelev.push_back(tr->val);
if(tr->left!=NULL)que.push(tr->left);
if(tr->right!=NULL)que.push(tr->right);

}else{
res.push_back(onelev);
onelev.clear();

if(que.empty())break;
que.push(NULL);
}
}
reverse(res.begin(),res.end());
return res;
}
};