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LeetCode: Binary Tree Level Order Traversal
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 };
LeetCode: Binary Tree Level Order Traversal
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