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]]

这一题和102题的做法是一样的，只是把上一题结果翻转一下就可以了。代码如下，注释内为递归版本。

Code:

/** * 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;        levelOrderBottom(root, 1, result);        reverse(result.begin(), result.end());        return result;    }    void levelOrderBottom(TreeNode *root, int level, vector<vector<int> > &result) {        if (!root) {            return;        }        if (result.size() < level) {            result.push_back(vector<int>());        }        result[level - 1].push_back(root -> val);        levelOrderBottom(root -> left, level + 1, result);        levelOrderBottom(root -> right, level + 1, result);    }*/    vector<vector<int>> levelOrderBottom(TreeNode* root) {        vector<vector<int>> result;        queue<TreeNode *> current, next;        if (!root) {            return result;        } else {            current.push(root);        }        TreeNode *pointer = root;        while (!current.empty()) {            vector<int> temp;            while (!current.empty()) {                pointer = current.front();                current.pop();                temp.push_back(pointer -> val);                if (pointer -> left) {                    next.push(pointer -> left);                }                if (pointer -> right) {                    next.push(pointer -> right);                }            }            result.push_back(temp);            swap(current, next);        }        reverse(result.begin(), result.end());        return result;    }};