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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
从底部层序遍历其实还是从顶部开始遍历，只不过最后存储的方式有所改变，可以参见我之前的博文 http://www.cnblogs.com/grandyang/p/4051321.html。 代码如下：
解法一：
class Solution { public: vector<vector<int> > levelOrderBottom(TreeNode *root) { vector<vector<int> > res; if (root == NULL) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { vector<int> oneLevel; int size = q.size(); for (int i = 0; i < size; ++i) { TreeNode *node = q.front(); q.pop(); oneLevel.push_back(node->val); if (node->left) q.push(node->left); if (node->right) q.push(node->right); } res.insert(res.begin(), oneLevel); } return res; } };
下面我们来看递归的解法，核心就在于我们需要一个二维数组，和一个变量level，当level递归到上一层的个数，我们新建一个空层，继续往里面加数字，参见代码如下：
解法二：
class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode* root) { vector<vector<int> > res; levelorder(root, 0, res); return vector<vector<int> > (res.rbegin(), res.rend()); } void levelorder(TreeNode *root, int level, vector<vector<int> > &res) { if (!root) return; if (res.size() == level) res.push_back({}); res[level].push_back(root->val); if (root->left) levelorder(root->left, level + 1, res); if (root->right) levelorder(root->right, level + 1, res); } };
本文转自博客园Grandyang的博客，原文链接：[LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二
，如需转载请自行联系原博主。
