Binary Tree Preorder Traversal
最后更新于:2022-04-01 22:56:28
**一. 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?
**二. 题目分析**
可使用递归解法,而只要是能用递归,也就是说能用栈来还原递归过程,因此方法不止一种。
使用栈来实现二叉树的遍历:首先在stack中压入当前的root,由于是前序遍历,故树是按照先根,然后左子树和后右子树进行访问,故pop取出一个结点,将它的value加入访问序列。之后压入它的右子树和左子树。直到stack为空。
**三. 示例代码**
递归解法:
~~~
#include
#include
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL){}
};
class Solution
{
public:
void preorder(TreeNode *root, vector &k)
{
if (root != NULL)
{
k.push_back(root->val);
preorder(root->left, k);
preorder(root->right, k);
}
}
vector preorderTraversal(TreeNode *root)
{
vector temp;
preorder(root, temp);
return temp;
}
};
~~~
非递归解法(stack实现):
~~~
#include
#include
#include
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL){}
};
class Solution {
public:
vector preorderTraversal(TreeNode *root) {
vector temp;
temp.clear();
stack k;
if (root == NULL)
return temp;
k.push(root);
while (!k.empty())
{
TreeNode *node = k.top();
k.pop();
temp.push_back(node->val);
if (node->right != NULL)
k.push(node->right);
if (node->left != NULL)
k.push(node->left);
}
return temp;
}
};
~~~
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