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LeetCode103 Binary Tree Zigzag Level Order Traversal
Given a binary tree, return the zigzag level order traversal of its nodes‘ values. (ie, from left to right, then right to left for the next level and alternate between).(Medium)
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3 / 9 20 / 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
分析:
层序遍历还要交叉输出。所以先采用队列进行层序编译(存放一层在队列中,每次循环先读这一队列的长度,然后以此把他们的值存在vector里,并且把存在的左右孩子压入队列)。
因为要交叉输出,所以采用一个flag,每次循环后flag正负号改变。负号时倒序输出。
代码:
1 /** 2 * Definition for a binary tree node. 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>> zigzagLevelOrder(TreeNode* root) { 13 vector<vector<int>> result; 14 if (root == nullptr) { 15 return result; 16 } 17 queue<TreeNode* >que; 18 que.push(root); 19 int flag = 1; 20 while(!que.empty()) { 21 int sz = que.size(); 22 vector<int> temp; 23 for (int i = 0; i < sz; ++i) { 24 TreeNode* cur = que.front(); 25 que.pop(); 26 temp.push_back(cur -> val); 27 if (cur -> left != nullptr) { 28 que.push(cur -> left); 29 } 30 if (cur -> right != nullptr) { 31 que.push(cur -> right); 32 } 33 } 34 if (flag == -1) { 35 reverse(temp.begin(), temp.end()); 36 result.push_back(temp); 37 flag = 1; 38 } 39 else { 40 result.push_back(temp); 41 flag = -1; 42 } 43 44 } 45 return result; 46 } 47 };
LeetCode103 Binary Tree Zigzag Level Order Traversal
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