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] ]
思路：用空指针标记一层结束，用bool变量标记访问的是奇数层还是偶数层。
时间复杂度O(n),空间复杂度O(1)
相关题目：《剑指offer》面试题61
1 /** 2 * Definition for binary tree 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> > ret; 14 if (root == nullptr) return ret; 15 16 vector<int> level; 17 queue<TreeNode *> q; 18 q.push(root); 19 q.push(0); 20 bool odd_even = true; 21 while (!q.empty()) { 22 TreeNode *p = q.front(); 23 q.pop(); 24 if (p) { 25 level.push_back(p->val); 26 if (p->left) 27 q.push(p->left); 28 if (p->right) 29 q.push(p->right); 30 } else { 31 if (!odd_even) { 32 reverse(level.begin(), level.end()); 33 } 34 35 odd_even = !odd_even; 36 ret.push_back(level); 37 level.clear(); 38 if (!q.empty()) 39 q.push(0); 40 41 } 42 } 43 44 return ret; 45 } 46 };
思路二：用两个栈。一个记录当前层节点，一个记录下一层节点。
1 /** 2 * Definition for binary tree 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> > ret; 14 if (root == nullptr) return ret; 15 16 vector<int> level; 17 stack<TreeNode *> current,next; 18 bool odd_even = true; 19 20 current.push(root); 21 while (!current.empty()) { 22 while (!next.empty()) { 23 TreeNode *p = current.top(); 24 current.pop(); 25 level.push_back(p->val); 26 27 if (odd_even) { 28 if (p->left) 29 next.push(p->left); 30 if (p->right) 31 next.push(p->right); 32 } else { 33 if (p->right) 34 next.push(p->right); 35 if (p->left) 36 next.push(p->left); 37 } 38 } 39 40 ret.push_back(level); 41 level.clear(); 42 swap(current, next); 43 odd_even = !odd_even; 44 45 } 46 47 return ret; 48 } 49 };