标签：span result vector val pop eve public while tree
思路一：使用非递归方式，当时还是需要注意每当一个算法写完后，检查是否处理到根节点或者list为空的情况
/** * 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; if(root == nullptr) return result; stack<TreeNode *> s; s.push(root); while(!s.empty()) { vector<TreeNode *> tmp; vector<int> _result; while(!s.empty()) { _result.push_back(s.top()->val); tmp.push_back(s.top()); s.pop(); } result.push_back(_result); for(int i=tmp.size()-1; i >= 0; --i) { if(tmp[i]->right != nullptr) s.push(tmp[i]->right); if(tmp[i]->left != nullptr) s.push(tmp[i]->left); } } return result; } };
思路二：采用递归的方式，这里使用一个额外参数记录了层次遍历的深度
/** * 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; tranverse(root, 1, result); return result; } void tranverse(TreeNode *root, int level, vector<vector<int>> &result) { if(root == nullptr) return; if(level > result.size()) result.push_back(vector<int>()); result[level-1].push_back(root->val); if(root->left != nullptr) tranverse(root->left, level+1, result); if(root->right != nullptr) tranverse(root->right, level+1, result); } };
Binary Tree Level Order Traversal
标签：span result vector val pop eve public while tree
原文地址：http://www.cnblogs.com/chengyuz/p/6658596.html
