LeetCode: Binary Tree Level Order Traversal
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] ]
地址：https://oj.leetcode.com/problems/binary-tree-level-order-traversal/
算法：BFS一颗树，然后分别输出每一层的节点值。在BFS遍历过程中，标识出每一层的结束节点即可。第一层的结束节点即为根，根据第一层的结束节点就可以得到第二层的结束节点，依次类推就可以得到各层的结束节点。代码：
1 /**
2 * Definition for binary tree
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 if(!root) return vector<vector<int> >();
14 queue<TreeNode *> que;
15 que.push(root);
16 int push_num = 1;
17 int pop_num = 0;
18 vector<vector<int> > result;
19 vector<int> temp;
20 int last = 1;
21 while(!que.empty()){
22 TreeNode *pop_node = que.front();
23 temp.push_back(pop_node->val);
24 que.pop();
25 ++pop_num;
26 if(pop_node->left){
27 que.push(pop_node->left);
28 ++push_num;
29 }
30 if(pop_node->right){
31 que.push(pop_node->right);
32 ++push_num;
33 }
34 if(pop_num == last){
35 result.push_back(temp);
36 temp.clear();
37 last = push_num;
38 }
39 }
40 return result;
41 }
42 };
第二题：
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] ]
地址：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/
算法：这一题要求先输出最后一层的节点。在上一题的基础上在吧结果转置一下，这样的方法会不会很不高端啊？代码：
1 /**
2 * Definition for binary tree
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 if(!root) return vector<vector<int> >();
14 queue<TreeNode *> que;
15 que.push(root);
16 int push_num = 1;
17 int pop_num = 0;
18 vector<vector<int> > result;
19 vector<int> temp;
20 int last = 1;
21 while(!que.empty()){
22 TreeNode *pop_node = que.front();
23 temp.push_back(pop_node->val);
24 que.pop();
25 ++pop_num;
26 if(pop_node->left){
27 que.push(pop_node->left);
28 ++push_num;
29 }
30 if(pop_node->right){
31 que.push(pop_node->right);
32 ++push_num;
33 }
34 if(pop_num == last){
35 result.push_back(temp);
36 temp.clear();
37 last = push_num;
38 }
39 }
40 temp.clear();
41 int len = result.size();
42 int half_len = len / 2;
43 for(int i = 0; i < half_len; ++i){
44 temp = result[i];
45 result[i] = result[len - i - 1];
46 result[len - i - 1] = temp;
47 }
48 return result;
49 }
50 };
