[Leetcode] 107. Binary Tree Level Order Traversal II 解题报告

题目：

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]]

思路：

这道题目和Leetcode 102的唯一区别就是最后需要把结果翻转一下，具体请参加我的Leetcode 102解题报告。

代码：

/** * 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) {        if (root == NULL) {            return {};        }        vector<vector<int>> ret;        queue<TreeNode*> q;        q.push(root);        q.push(NULL);       // separator        vector<int> line;        while (!q.empty()) {            TreeNode* node = q.front();            q.pop();            if (node == NULL) {                ret.push_back(line);                line.clear();                if (!q.empty()) {                    q.push(NULL);                }            }            else {                line.push_back(node->val);                if (node->left) {                    q.push(node->left);                }                if (node->right) {                    q.push(node->right);                }            }        }        reverse(ret.begin(), ret.end());    // the only difference with Leetcode 102        return ret;    }};