[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] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
1 BFS迭代，借用之前min depth和maxdepth of bianry tree 的框架
1 struct newNode{ 2 TreeNode * node; 3 int dep; // current depth 4 }; 5 6 class Solution { 7 public: 8 vector< vector<int> > levelOrder(TreeNode *root) { 9 vector<vector<int> > result; 10 if(root == NULL) return result; 11 12 vector<int> levelRes; 13 queue<newNode*> que; 14 int curDep ; 15 16 newNode *p = new newNode; 17 p->node = root; 18 curDep = p->dep = 1; 19 que.push(p); 20 21 while( !que.empty()) 22 { 23 p= que.front(); 24 que.pop(); 25 26 TreeNode *pNode = p->node; 27 int dep = p->dep; 28 29 if(curDep == dep) 30 { 31 levelRes.push_back(pNode->val); 32 } 33 else 34 { 35 result.push_back(levelRes); 36 levelRes.clear(); 37 levelRes.push_back(pNode->val); 38 curDep++; 39 } 40 41 if(pNode->left != NULL) 42 { 43 newNode *tmp = new newNode; 44 tmp->node = pNode->left; 45 tmp->dep = dep + 1; 46 que.push(tmp); 47 } 48 49 if(pNode->right != NULL) 50 { 51 newNode *tmp = new newNode; 52 tmp->node = pNode->right; 53 tmp->dep = dep + 1; 54 que.push(tmp); 55 } 56 57 } 58 59 // add the last level 60 result.push_back(levelRes); 61 62 return result; 63 } 64 65 };
2 DFS 也可以解决，就是在DFS中每次要传入一个dep，然后根据dep将数据加入到对应的vector中。
DFS 递归（使用前序遍历框架）
1 class Solution { 2 private: 3 vector<vector<int> > m_result; 4 public: 5 void solve(int dep, TreeNode *root) 6 { 7 if (root == NULL) 8 return; 9 10 if (m_result.size() > dep) 11 { 12 m_result[dep].push_back(root->val); 13 } 14 else 15 { 16 vector<int> tmp; 17 tmp.push_back(root->val); 18 m_result.push_back(tmp); 19 } 20 21 solve(dep + 1, root->left); 22 solve(dep + 1, root->right); 23 } 24 25 vector<vector<int> > levelOrder(TreeNode *root) { 26 // Start typing your C/C++ solution below 27 // DO NOT write int main() function 28 m_result.clear(); 29 solve(0, root); 30 31 return m_result; 32 } 33 };
DFS 迭代
1 struct newNode{ 2 TreeNode * node; 3 int dep; // current depth 4 }; 5 6 class Solution { 7 private: 8 //vector<vector<int> > retsult; 9 public: 10 vector< vector<int> > levelOrder(TreeNode *root) { 11 12 vector<vector<int> > result; 13 if(root == NULL) return result; 14 15 vector<int> levelRes; 16 stack<newNode*> s; 17 18 newNode *p = new newNode; 19 p->node = root; 20 p->dep = 0; 21 s.push(p); 22 23 while (!s.empty()) { 24 p = s.top(); 25 s.pop(); 26 27 TreeNode *pNode = p->node; 28 int dep = p->dep; 29 30 if( dep + 1 > result.size() ) // new a vector 31 { 32 vector<int> tmp; 33 tmp.push_back(pNode->val); 34 result.push_back(tmp); 35 } 36 else // vector exists 37 { 38 result[dep].push_back(pNode->val); 39 } 40 //注意先压入右子树，再压入左子树，弹出的时候先弹出左子树，在弹出右子树，实现前序遍历，Root->left->right 41 if(pNode->right != NULL) 42 { 43 newNode *tmp = new newNode; 44 tmp->node = pNode->right; 45 tmp->dep = dep + 1; 46 s.push(tmp); 47 } 48 49 if(pNode->left != NULL) 50 { 51 newNode *tmp = new newNode; 52 tmp->node = pNode->left; 53 tmp->dep = dep + 1; 54 s.push(tmp); 55 } 56 } 57 return result; 58 } 59 };
3 双queue法
class Solution { public: vector<vector<int> > levelOrder(TreeNode *root) { queue<TreeNode*> q1; queue<TreeNode*> q2; vector<vector<int> > res; if(root != NULL) { q1.push(root); res.push_back(vector<int> ()); } int depth = 0; while(!q1.empty()) { TreeNode * p = q1.front(); q1.pop(); res[depth].push_back(p->val); if(p->left) q2.push(p->left); if(p->right) q2.push(p->right); if(q1.empty() && !q2.empty()) { swap(q1, q2); depth ++; res.push_back(vector<int> ()); } } return res; } };
