LeetCode: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]
] 本文地址
分析：首先肯定的是要对树进行层序遍历，但是相邻两层的元素遍历顺序是相反的，因此传统的非递归遍历方法用一个队列肯定是无法实现。我们用两个栈来实现，同一层元素都在同一个栈中，相邻的两层元素放在不同的栈中，比如第一层元素放在第一个栈，那么第二层就放在第二个栈，第三层就放在第三个栈......如果某一层元素是从左往右遍历的，那么这层元素的孩子节点入栈顺序就是先左孩子后右孩子，相反如果某一层元素是从右往左遍历的，入栈顺序就是先右孩子后左孩子。
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 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 vector<vector<int> >res; 16 if(root == NULL)return res; 17 stack<TreeNode*> S,S2; 18 S.push(root); 19 bool isS = true; 20 vector<int>tmpres; 21 while(S.empty() == false || S2.empty() == false) 22 { 23 TreeNode *p; 24 if(isS) 25 { 26 27 p = S.top(); 28 S.pop(); 29 tmpres.push_back(p->val); 30 if(p->left)S2.push(p->left); 31 if(p->right)S2.push(p->right); 32 if(S.empty() == true) 33 { 34 res.push_back(tmpres); 35 tmpres.clear(); 36 isS = false; 37 } 38 } 39 else 40 { 41 p = S2.top(); 42 S2.pop(); 43 tmpres.push_back(p->val); 44 if(p->right)S.push(p->right); 45 if(p->left)S.push(p->left); 46 if(S2.empty() == true) 47 { 48 res.push_back(tmpres); 49 tmpres.clear(); 50 isS = true; 51 } 52 } 53 } 54 return res; 55 } 56 };
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