144. 二叉树的前序遍历(非递归实现)
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?
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
代码实现
C++ 实现
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root)
{
vector<int> ret;
//利用程序中的栈,模拟系统堆栈的功能
stack<TreeNode*> s;
//路径上的当前的节点
TreeNode *p=root;
while(p!=NULL||!s.empty())
{
//按照先序遍历的顺序,栈中存储的是先序遍历经过的路径上遇到
//的节点,根据先序遍历的特点,这个循环中已经将根节点,左子节点访问过了
while(p!=NULL)
{
ret.push_back(p->val);
s.push(p);
p=p->left;
}
//之所以要用栈保存访问过的路径上经过的节点,是因为我们只能由父节点得到左、右
//子节点,所以经过的路径上的节点还有用,不能丢弃,我们暂时保存在栈中,而根据先
//序遍历的特点,我们需要从经过的路径的下面到上面进行回溯。这就需要栈的回退的特点
if(!s.empty())
{
p=s.top();
s.pop();
p=p->right;
}
}
return ret;
}
}
node.h:
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
pre_order.cpp:
#include "node.h"
#include <vector>
#include <stack>
using namespace std;
vector<int> preorderTraversal(TreeNode *root)
{
if (root == nullptr)
return {};
vector<int> ret;
stack<TreeNode *> stk;
TreeNode *p = root;
while (true)
{
while (p != nullptr)
{
ret.push_back(p->val);
stk.push(p);
p = p->left;
}
if (!stk.empty())
{
TreeNode *temp = stk.top();
stk.pop();
p = temp->right;
}
else
break;
}
return ret;
}
java实现
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
Deque<TreeNode> stk = new LinkedList<TreeNode>();
TreeNode p = root;
while (true) {
while (p != null) {
ret.add(p.val);
stk.push(p);
p = p.left;
}
if (!stk.isEmpty()) {
p = stk.peek();
stk.pop();
p = p.right;
} else
break;
}
return ret;
}
}