[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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

这题和之前那题差不多，加个reverse函数就可以。

/** * 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) {        vector<vector<int>> result;        if (root == nullptr)            return result;                queue<TreeNode*> Q;        Q.push(root);        while (!Q.empty()) {            int size = Q.size();            vector<int> level;                        for (int i = 0; i != size; ++i) {                TreeNode* head = Q.front();                Q.pop();                level.push_back(head->val);                if (head->left != nullptr)                    Q.push(head->left);                if (head->right != nullptr)                    Q.push(head->right);            }            result.push_back(level);        }        reverse(result.begin(), result.end());        return result;    }};