107. Binary Tree Level Order Traversal II

题目：Binary Tree Level Order Traversal II

原题描述

原题链接：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/#/description
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],

return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]

从尾到头层序遍历二叉树。
例：二叉树如图，则结果为[ [15,7], [9,20], [3] ]。

思路

http://blog.csdn.net/gcs6564157/article/details/62451157
和这题一样，只需要最后把数组反转一下就可以了。

参考代码

/** * 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) {        vector<vector<int>> ans;        if(!root) return ans;        vector<int> temp;        queue<TreeNode*> q;        q.push(root);        TreeNode* firstLeft = NULL;        bool findFirst = false;        while(!q.empty()) {            TreeNode* front = q.front();            q.pop();            if(front == firstLeft) {                ans.push_back(temp);                temp.clear();                firstLeft = NULL;                findFirst = false;            }            temp.push_back(front->val);            if(front->left){                if(!findFirst) {                    firstLeft = front->left;                    findFirst = true;                }                q.push(front->left);            }            if(front->right) {                if(!findFirst) {                    firstLeft = front->right;                    findFirst = true;                }                q.push(front->right);            }        }        ans.push_back(temp);        return ans;     }};