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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?
题目解析:
将一颗二叉树以前序遍历的顺序存入一个vector中。先将root根节点存入vector中,再递归左子树和右子树。和前序遍历不同之处在于需要一个函数以传引用的方式将vector传入。
AC代码:
//递归版
/**
* 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> v;
if(root == NULL)
return v;
PreOrder(root,v);
return v;
}
void PreOrder(TreeNode* root,vector<int>& res)
{
if(root == NULL)
return;
res.push_back(root->val);
PreOrder(root->left,res);
PreOrder(root->right,res);
}
};
//非递归
/**
* 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> v;
if(root == NULL)
return v;
stack<TreeNode*> s;
s.push(root);
while(!s.empty())
{
TreeNode* top = s.top();
s.pop();
v.push_back(top->val);
if(top->right)
s.push(top->right);
if(top->left)
s.push(top->left);
}
return v;
}
};
(*^▽^*)