103. Binary Tree Zigzag Level Order Traversal

2019-11-10 21:45:54

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跟上一题一样，就最后换了vector的顺序。。。下次就应该在queue中换顺序吧

/** * 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>> zigzagLevelOrder(TreeNode* root) { queue<TreeNode*>qu; vector<vector<int>>ve; if(root == NULL) return ve; vector<int>vec; int num = 1; qu.push(root); while(!qu.empty()){ TreeNode* t; vec.clear(); int nn = num; num = 0; for(int i = 1; i <= nn; ++ i){ t = qu.front(); qu.pop(); vec.push_back(t -> val); if(t -> left != NULL){ num++; qu.push(t -> left); } if(t -> right != NULL){ num++; qu.push(t -> right); } } ve.push_back(vec); } for(int i = 0; i < ve.size(); ++ i){ if(i % 2 == 0) continue; for(int j = 0; j < ve[i].size() / 2; ++ j){ int n = ve[i].size(); int t; t = ve[i][n - 1 - j]; ve[i][n - 1 - j] = ve[i][j]; ve[i][j] = t; } } return ve; }};

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