【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：层次遍历二叉树，然后再进行逆序操作；

class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root)
{
vector<vector<int>> res;

if (root == NULL)
{
return res;
}

queue<TreeNode*> q;
q.push(root);

while (q.size()>0)
{
vector<int> level;
for (int i = 0, n = q.size(); i<n; i++)
{
TreeNode* tmp = q.front();
q.pop();

level.push_back(tmp->val);

if (tmp->left)
{
q.push(tmp->left);
}
if (tmp->right)
{
q.push(tmp->right);
}
}

res.push_back(level);
}

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

return res;
}
};