leetcode 102 Binary Tree Level Order Traversal
2015-07-14 22:53
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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
return its level order traversal as:
confused what
> read more on how binary tree is serialized on OJ.
我的解决方案,非常传统的两个队列的解决方案:
一个栈似乎也行:
递归的解决方案:
逆序排列把return 改一下就好了: return
vector<vector<int> > (result.rbegin(), result.rend());
}
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.
我的解决方案,非常传统的两个队列的解决方案:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> result; queue<TreeNode*> q; if (root == NULL) { return result; } q.push(root); vector<int> le_temp; while(!q.empty()) { le_temp.clear(); queue<TreeNode*> level; int size = q.size(); for(int i = 0; i < size; ++i) { TreeNode* temp = q.front(); q.pop(); if(temp->left) { level.push(temp->left); } if(temp->right) { level.push(temp->right); } le_temp.push_back(temp->val); } while(!level.empty()) { q.push(level.front()); level.pop(); } result.push_back(le_temp); } return result; } };
一个栈似乎也行:
vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int> > result; if (!root) return result; queue<TreeNode*> q; q.push(root); q.push(NULL); vector<int> cur_vec; while(!q.empty()) { TreeNode* t = q.front(); q.pop(); if (t==NULL) { result.push_back(cur_vec); cur_vec.resize(0); if (q.size() > 0) { q.push(NULL); } } else { cur_vec.push_back(t->val); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } } return result; }
递归的解决方案:
class Solution { public: vector<vector<int>> result; void buildVector(TreeNode* root, int depth) { if(root == NULL)return ; if(result.size() == depth) { result.push_back(vector<int>()); } result[depth].push_back(root->val); buildVector(root->left,depth + 1); buildVector(root->right, depth + 1); } vector<vector<int>> levelOrder(TreeNode* root) { buildVector(root,0); return result; } };
逆序排列把return 改一下就好了: return
vector<vector<int> > (result.rbegin(), result.rend());
}
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