Given a binary tree, return the preorder traversal of its nodes’ values.
For example:
Given binary tree {1,#,2,3},
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
递归实现代码
/*****************************************************************
* @Author : 楚兴
* @Date : 2015/2/23 18:27
* @Status : Accepted
* @Runtime : 4 ms
******************************************************************/
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL){}
};
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
helper(root, result);
return result;
}
void helper(TreeNode *root, vector<int>& num)
{
if (root)
{
num.push_back(root->val);
helper(root->left, num);
helper(root->right, num);
}
}
};
非递归实现代码1
/*****************************************************************
* @Author : 楚兴
* @Date : 2015/2/24 00:54
* @Status : Accepted
* @Runtime : 6 ms
******************************************************************/
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode*> nodes;
TreeNode *p = root;
if (p)
{
nodes.push(p);
while (!nodes.empty())
{
p = nodes.top();
nodes.pop();
result.push_back(p->val);
if (p->right)
{
nodes.push(p->right);
}
if (p->left)
{
nodes.push(p->left);
}
}
}
return result;
}
};
非递归实现代码2
/*****************************************************************
* @Author : 楚兴
* @Date : 2015/2/24 01:06
* @Status : Accepted
* @Runtime : 4 ms
******************************************************************/
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode*> nodes;
TreeNode *p = root;
while (p != NULL || !nodes.empty())
{
if (p != NULL)
{
while(p != NULL)
{
result.push_back(p->val);
nodes.push(p);
p = p->left;
}
}
else
{
p = nodes.top()->right;
nodes.pop();
}
}
return result;
}
};
后序遍历的实现见博文Binary Tree Postorder Traversal
时间: 2024-11-10 01:32:01