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] ]
用BFS然后根据层数来判断是否要反转列表
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 11 struct Node 12 { 13 TreeNode *node; 14 int level; 15 Node(){} 16 Node(TreeNode *n, int l):node(n), level(l){} 17 }; 18 19 class Solution { 20 private: 21 vector<vector<int> > ret; 22 public: 23 vector<vector<int> > zigzagLevelOrder(TreeNode *root) { 24 // Start typing your C/C++ solution below 25 // DO NOT write int main() function 26 ret.clear(); 27 28 if (root == NULL) 29 return ret; 30 31 queue<Node> q; 32 33 q.push(Node(root, 0)); 34 35 int curLevel = -1; 36 37 vector<int> a; 38 39 while(!q.empty()) 40 { 41 Node node = q.front(); 42 43 if (node.node->left) 44 q.push(Node(node.node->left, node.level + 1)); 45 46 if (node.node->right) 47 q.push(Node(node.node->right, node.level + 1)); 48 49 if (curLevel != node.level) 50 { 51 if (curLevel != -1) 52 { 53 if (curLevel % 2 == 1) 54 reverse(a.begin(), a.end()); 55 56 ret.push_back(a); 57 } 58 a.clear(); 59 curLevel = node.level; 60 } 61 a.push_back(node.node->val); 62 q.pop(); 63 } 64 65 if (curLevel % 2 == 1) 66 reverse(a.begin(), a.end()); 67 68 ret.push_back(a); 69 70 return ret; 71 } 72 };