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题目要求：
Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
这道题利用宽度优先搜索就可以了，具体程序（8ms）如下：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> levelOrder(TreeNode* root) { 13 vector<vector<int>> retVec; 14 if(!root) 15 return retVec; 16 17 retVec.push_back(vector<int>{root->val}); 18 queue<TreeNode *> que; 19 que.push(root); 20 while(true) 21 { 22 vector<int> vec; 23 queue<TreeNode *> q; 24 while(!que.empty()) 25 { 26 TreeNode *tree = que.front(); 27 que.pop(); 28 if(tree->left) 29 { 30 vec.push_back((tree->left)->val); 31 q.push(tree->left); 32 } 33 if(tree->right) 34 { 35 vec.push_back((tree->right)->val); 36 q.push(tree->right); 37 } 38 } 39 40 if(!q.empty()) 41 { 42 que = q; 43 retVec.push_back(vec); 44 } 45 else 46 break; 47 } 48 49 return retVec; 50 } 51 };
题目要求：
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] ]
这道题基本更上边的题目一样，我们只需要将上题的结果retVec最后再反转一下就可以了，即添加如下一行即可：
1 reverse(retVec.begin(), retVec.end());
这样的程序只要8ms，但下边基本一样的程序却要64ms：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> levelOrderBottom(TreeNode* root) { 13 vector<vector<int>> retVec; 14 if(!root) 15 return retVec; 16 17 retVec.insert(retVec.begin(), vector<int>{root->val}); 18 queue<TreeNode *> que; 19 que.push(root); 20 while(true) 21 { 22 vector<int> vec; 23 queue<TreeNode *> q; 24 while(!que.empty()) 25 { 26 TreeNode *tree = que.front(); 27 que.pop(); 28 if(tree->left) 29 { 30 vec.push_back((tree->left)->val); 31 q.push(tree->left); 32 } 33 if(tree->right) 34 { 35 vec.push_back((tree->right)->val); 36 q.push(tree->right); 37 } 38 } 39 40 if(!q.empty()) 41 { 42 que = q; 43 retVec.insert(retVec.begin(), vec); 44 } 45 else 46 break; 47 } 48 49 return retVec; 50 } 51 };
LeetCode之“树”：Binary Tree Level Order Traversal && Binary Tree Level Order Traversal II
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原文地址：http://www.cnblogs.com/xiehongfeng100/p/4631447.html
