leetcode--107--二叉树的层次遍历 II

题目：
给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

例如：
给定二叉树 [3,9,20,null,null,15,7],

3

/
9 20
/
15 7
返回其自底向上的层次遍历为：

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

链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii

思路：
1、模板题，层次遍历后前后翻转一下即可

Python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):

    def __init__(self):
        self.ret = []

    def bfs(self, root):
        if not root:
            return []
        deque = [root]

        while deque:
            temp = []
            nex = []
            for item in deque:
                temp.append(item.val)
                if item.left:
                    nex.append(item.left)
                if item.right:
                    nex.append(item.right)
            deque = nex
            self.ret.append(temp)
        return self.ret[::-1]

    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        return self.bfs(root)

C++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>> ret;

    vector<vector<int>> bfs(TreeNode* root){
        if(root == nullptr){
            return ret;
        }
        vector<TreeNode* > deque;
        deque.push_back(root);
        
        while(deque.size()>0){
            vector<TreeNode* > nex;
            vector<int> temp;
            for (auto item : deque){
                temp.push_back(item->val);
                if (item->left){
                    nex.push_back(item->left);
                }
                if (item->right){
                    nex.push_back(item->right);
                }
            }
            deque.swap(nex);
            ret.push_back(temp);
        }
        reverse(ret.begin(), ret.end());
        return ret;
    }

    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        return bfs(root);
    }
};