【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;    }};