前言
【LeetCode 题解】系列传送门: http://www.cnblogs.com/double-win/category/573499.html
1.题目描述
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?
2. 题意
先序遍历二叉树,递归的思路是普通的,能否用迭代呢?
3. 思路
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);
}
};
4.相关题目
(1)二叉树的中序遍历:
(2)二叉树的后序遍历:
(3) 二叉树系列文章:
作者:Double_Win 出处: http://www.cnblogs.com/double-win/p/3896010.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~ |