题目:
Given a binary tree, return the preorder traversal of its nodes' values.
给定一个二叉树,返回前序遍历的结果。
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?
思路:
非递归遍历时的思路是先输出结点值,再入栈,然后遍历左子树。退栈时,遍历栈顶结点的右子树。
详情参看:http://blog.youkuaiyun.com/u012243115/article/details/40615603 。
递归版代码:
/**
* Definition for binary tree
* 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;
if(root == NULL)
return result;
preorderTraversal(root,result);
return result;
}
void preorderTraversal(TreeNode *root , vector<int> &result)
{
if(root == NULL)
return;
result.push_back(root->val);
preorderTraversal(root->left,result);
preorderTraversal(root->right,result);
}
};
非递归版代码一:
/**
* Definition for binary tree
* 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;
stack<TreeNode *> stk;
TreeNode *p = root;
while(p != NULL || !stk.empty())
{
if(p != NULL)
{
result.push_back(p->val);
stk.push(p);
p = p->left;
}
else
{
p = stk.top();
stk.pop();
p = p->right;
}
}
return result;
}
};
非递归版代码二:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
//思路是根结点先入栈,然后根结点出栈,输出根结点,右孩子入栈(如果有的话)。左孩子入栈(如果有的话)。
vector<int> preorderTraversal(TreeNode *root)
{
vector<int> result;
stack<TreeNode *> stk;
if(root == NULL)
return result;
TreeNode *p = root;
stk.push(p);
while(!stk.empty())
{
p = stk.top();
result.push_back(p->val);
stk.pop();
if(p->right)
stk.push(p->right);
if(p->left)
stk.push(p->left);
}
return result;
}
};
使用链表实现栈的效果:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
//使用链表实现栈的效果,操作都一样,只是出栈方式不一样
vector<int> preorderTraversal(TreeNode *root)
{
vector<int> result;
list<TreeNode *> lst;
if(root == NULL)
return result;
TreeNode *p = root;
lst.push_front(p);//插入链表头,相当于入栈
while(!lst.empty())
{
p = lst.front();//取链表头
result.push_back(p->val);
lst.erase(lst.begin());//删除链表头
if(p->right)
lst.push_front(p->right);
if(p->left)
lst.push_front(p->left);
}
return result;
}
};