[leetcode] 107. Binary Tree Level Order Traversal II

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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;}
};

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