【leetcode】107. 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]
]

/**
* Definition for a binary tree node.
* 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>> vec;
if (!root) {
return vec;
}

stack<vector<int>> st;

q.push(root);
q.push(NULL);

vector<int> tmp;
while (!q.empty()) {
TreeNode *node = q.front();
q.pop();

if (node == NULL) {
st.push(tmp);
tmp.clear();
if (q.empty()) {
break;
}
q.push(NULL);
continue;
}

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

while (!st.empty()) {
vec.push_back(st.top());
st.pop();
}

return vec;
}
};

[/code]