Breadth-first Search -- Leetcode problem107. Binary Tree Level Order Traversal II

• 描述：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]
]

• 分析：层序遍历二叉树，将遍历结果从底层到高层输出
• 思路一：直接用reverse函数对Binary Tree Level Order Traversal中的vector结果进行反转操作。

class Solution {public:    vector<vector<int>> levelOrderBottom(TreeNode* root) {    vector<vector<int>> my_vec;    vector<int> result;    if (!root) return my_vec;    queue<TreeNode*> my_queue;    my_queue.push(root);    while (!my_queue.empty()) {        int n = my_queue.size();        result.clear();        for (int i = 0; i < n; i ++) {            TreeNode* temp = my_queue.front();            my_queue.pop();            result.push_back(temp -> val);            if (temp -> left) my_queue.push(temp -> left);            if (temp -> right) my_queue.push(temp -> right);        }        my_vec.push_back(result);    }    reverse(my_vec.begin(), my_vec.end());    return my_vec;}};