【Leetcode】【Medium】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).
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] ]
解题思路：
按层遍历，建立两个栈，一个保存当前结点，一个按照Z型保存下一层结点。
使用一个bool变量来标注每次读取子节点的方向；
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> > ret; 14 stack<TreeNode*> cur_layer; 15 stack<TreeNode*> next_layer; 16 cur_layer.push(root); 17 bool dirt = false; 18 19 if (!root) 20 return ret; 21 22 while (!cur_layer.empty()) { 23 vector<int> layer_val; 24 while (!cur_layer.empty()) { 25 TreeNode* node = cur_layer.top(); 26 cur_layer.pop(); 27 layer_val.push_back(node->val); 28 if (dirt) { 29 if (node->right) 30 next_layer.push(node->right); 31 if (node->left) 32 next_layer.push(node->left); 33 } else { 34 if (node->left) 35 next_layer.push(node->left); 36 if (node->right) 37 next_layer.push(node->right); 38 } 39 } 40 ret.push_back(layer_val); 41 swap(cur_layer, next_layer); 42 dirt = !dirt; 43 } 44 return ret; 45 } 46 };
