标签：ntc empty evel color 不能 erb tom des res
题目：
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).
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] ]
题解：
迭代法，利用队列，不能用栈。
Solution 1 ()
class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode *root) { vector<vector<int>> res; if (root == nullptr) { return res; } vector<int> v; queue<TreeNode*> q; q.push(root); q.push(nullptr); while (!q.empty()) { TreeNode* cur = q.front(); q.pop(); if(cur == nullptr) { res.insert(res.begin(),v); v.clear(); if(!q.empty()) { q.push(nullptr); } continue; } v.push_back(cur->val); if(cur->left != nullptr) { q.push(cur->left); } if(cur->right != nullptr) { q.push(cur->right); } } return res; } };
Solution 1.1 ()
class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode *root) { vector<vector<int>> res; if (root == nullptr) { return res; } queue<TreeNode*> q; q.push(root); while (!q.empty()) { int size = q.size(); vector<int> v; for (int i = 0; i < size; ++i) { TreeNode* cur = q.front(); q.pop(); if(cur->left != nullptr) { q.push(cur->left); } if(cur->right != nullptr) { q.push(cur->right); } v.push_back(cur->val); } res.insert(res.begin(),v); } return res; } };
递归
Solution 2 ()
【Lintcode】070.Binary Tree Level Order Traversal II
标签：ntc empty evel color 不能 erb tom des res
原文地址：http://www.cnblogs.com/Atanisi/p/6906905.html
