跟1基本一样，就是用个stack就解决了，算一次过吧
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 void dfs(queue<TreeNode *> &pre, stack<vector<int>> &save) { 13 queue<TreeNode *> S; 14 vector<int> level; 15 while (!pre.empty()) { 16 TreeNode *tmp = pre.front(); 17 if (tmp->left) S.push(tmp->left); 18 if (tmp->right) S.push(tmp->right); 19 level.push_back(tmp->val); 20 pre.pop(); 21 } 22 save.push(level); 23 if (!S.empty()) { 24 pre = S; 25 dfs(pre, save); 26 } 27 } 28 vector<vector<int> > levelOrderBottom(TreeNode *root) { 29 // Start typing your C/C++ solution below 30 // DO NOT write int main() function 31 stack<vector<int>> save; 32 queue<TreeNode *> pre; 33 vector<vector<int>> ret; 34 if (!root) return ret; 35 pre.push(root); 36 dfs(pre, save); 37 while (!save.empty()) { 38 vector<int> tmp = save.top(); 39 ret.push_back(tmp); 40 save.pop(); 41 } 42 return ret; 43 } 44 };
后来写了个更好的代码
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 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 vector<vector<int> > ret; 16 if (!root) return ret; 17 vector<int> cur; 18 queue<TreeNode *> S, T; 19 S.push(root); 20 while (!S.empty()) { 21 cur.clear(); 22 while (!S.empty()) { 23 TreeNode *tmp = S.front(); 24 S.pop(); 25 cur.push_back(tmp->val); 26 if (tmp->left) T.push(tmp->left); 27 if (tmp->right) T.push(tmp->right); 28 } 29 ret.push_back(cur); 30 S.swap(T); 31 } 32 reverse(ret.begin(), ret.end()); 33 return ret; 34 } 35 };
再加个recursive的方法
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 private: 12 vector<vector<int> > res; 13 public: 14 void dfs(TreeNode *root, int d) { 15 if (!root) return; 16 if (d >= res.size()) res.push_back(vector<int>(0)); 17 res[d].push_back(root->val); 18 dfs(root->left, d+1); 19 dfs(root->right, d+1); 20 } 21 vector<vector<int> > levelOrderBottom(TreeNode *root) { 22 // IMPORTANT: Please reset any member data you declared, as 23 // the same Solution instance will be reused for each test case. 24 res.clear(); 25 dfs(root, 0); 26 for (int i = 0; i < res.size()/2; i++) swap(res[i], res[res.size()-1-i]); 27 return res; 28 } 29 };
C#:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public int val; 5 * public TreeNode left; 6 * public TreeNode right; 7 * public TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<List<int>> LevelOrderBottom(TreeNode root) { 12 List<List<int>> ans = new List<List<int>>(); 13 if (root == null) return ans; 14 List<int> tmp = new List<int>(); 15 Queue<TreeNode> que = new Queue<TreeNode>(); 16 que.Enqueue(root); 17 que.Enqueue(null); 18 while (true) 19 { 20 TreeNode peek = que.Peek(); 21 que.Dequeue(); 22 if (peek == null) 23 { 24 ans.Add(tmp); 25 if (que.Count == 0 || que.Peek() == null) break; 26 tmp = new List<int>(); 27 que.Enqueue(null); 28 } 29 else { 30 if (peek.left != null) que.Enqueue(peek.left); 31 if (peek.right != null) que.Enqueue(peek.right); 32 tmp.Add(peek.val); 33 } 34 } 35 ans.Reverse(); 36 return ans; 37 } 38 }