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).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
1 class Solution { 2 public: 3 void reverse(vector<int> &arr) { 4 for (int i = 0, j = arr.size() - 1; i < j; i++, j--) swap(arr[i], arr[j]); 5 } 6 7 vector<vector<int> > zigzagLevelOrder(TreeNode *root) { 8 vector<vector<int> > ans; 9 if (root == NULL) return ans; 10 vector<vector<TreeNode*> > layers(2); 11 vector<int> tmp; 12 int cur = 0, next = 1; 13 layers[cur].push_back(root); 14 15 while (!layers[cur].empty()) { 16 layers[next].clear(); 17 tmp.clear(); 18 for (auto node : layers[cur]) { 19 if (node->left) layers[next].push_back(node->left); 20 if (node->right) layers[next].push_back(node->right); 21 tmp.push_back(node->val); 22 } 23 if (cur == 1) reverse(tmp); 24 ans.push_back(tmp); 25 cur = !cur, next = !next; 26 } 27 28 return ans; 29 } 30 };
Leetcode | Binary Tree Zigzag Level Order Traversal
原文:http://www.cnblogs.com/linyx/p/4041405.html