[leetcode] 107. Binary Tree Level Order Traversal II

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]
]

我的代码

同上一题，简单的bfs。
但是打印的顺序是由下至上，所以用栈来保存中间结果，最后把结果倒出来给vector。

/** * 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) {        stack<vector<int>> s;        vector<vector<int>> rlt;        queue<TreeNode*> q;        if (root)        {            q.push(root);            while(!q.empty())            {                vector<int> lyrRlt;                int _siz=q.size();                while(_siz)                {                    TreeNode* tmp=q.front();                    lyrRlt.push_back(tmp->val);                    if (tmp->left) q.push(tmp->left);                    if (tmp->right) q.push(tmp->right);                    q.pop();                    _siz--;                }                s.push(lyrRlt);            }            while(!s.empty())            {                vector<int> temp=s.top();                rlt.push_back(temp);                s.pop();            }        }        return rlt;    }};