Binary Tree Zigzag Level Order Traversal -- leetcode
2015-05-06 16:23
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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
return its zigzag level order traversal as:
基本思路:
1. 使用广度优先搜索
2. 交替使用两个栈。一个表示当前层,一个表示下一层的节点。
3. 使用一个标志变量,表示该层中,先入左节点,还是先入右结点。
在leetcode上实际执行时间为4ms。
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] ]
基本思路:
1. 使用广度优先搜索
2. 交替使用两个栈。一个表示当前层,一个表示下一层的节点。
3. 使用一个标志变量,表示该层中,先入左节点,还是先入右结点。
在leetcode上实际执行时间为4ms。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector<vector<int> > ans; if (!root) return ans; stack<TreeNode *> *next = new stack<TreeNode *>(); stack<TreeNode *> *curr = new stack<TreeNode *>(); curr->push(root); bool flag = true; while (!curr->empty()) { ans.push_back(vector<int>()); ans.back().reserve(curr->size()); while (!curr->empty()) { root = curr->top(); curr->pop(); ans.back().push_back(root->val); if (flag) { if (root->left) next->push(root->left); if (root->right) next->push(root->right); } else { if (root->right) next->push(root->right); if (root->left) next->push(root->left); } } swap(next, curr); flag = !flag; } delete next; delete curr; return ans; } };
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