leetcode_c++:树: Binary Tree Level Order Traversal II (107)

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,null,null,15,7],

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7],
[9,20],
[3]
]

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

if (root == NULL)
return res;

// bfs
q.push(root);
q.push(NULL);    // NULL is the seperator of levels
while (!q.empty()) {
front = q.front();
q.pop();
if (front) {
tmpv.push_back(front->val);
if (front->left)
q.push(front->left);
if (front->right)
q.push(front->right);
} else if (!tmpv.empty()) {
res.push_back(tmpv);
tmpv.clear();
q.push(NULL);
}
}

// reverse the result
reverse(res.begin(), res.end());
return res;
}
};