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Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
题意
先序遍历二叉树，递归的思路是普通的，能否用迭代呢？
非递归思路：<借助stack>
vector<int> preorderTraversal(TreeNode *root) { stack<TreeNode* > st; vector<int> vi; vi.clear(); if(!root) return vi; st.push(root); while(!st.empty()){ TreeNode *tmp = st.top(); vi.push_back(tmp->val); st.pop(); if(tmp->right) st.push(tmp->right); if(tmp->left) st.push(tmp->left); } return vi; }
递归思路：
class Solution { private: vector<int> vi; public: vector<int> preorderTraversal(TreeNode *root) { vi.clear(); if(!root) return vi; preorder(root);return vi; } void preorder(TreeNode* root){ if(!root) return; vi.push_back(root->val); preorder(root->left); preorder(root->right); } };
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作者：Double_Win 出处: http://www.cnblogs.com/double-win/p/3895822.html 声明： 由于本人水平有限，文章在表述和代码方面如有不妥之处，欢迎批评指正~ |
[LeetCode 题解]: Binary Tree Preorder Traversal,布布扣,bubuko.com
[LeetCode 题解]: Binary Tree Preorder Traversal
标签：des style blog http color io strong for
原文地址：http://www.cnblogs.com/double-win/p/3895822.html
