代码思路
递归的还是很简单的。
题中强调是迭代。
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void search(TreeNode* root,vector<int> &v){
if(!root) return ;
v.push_back(root->val);
search(root->left,v);
search(root->right,v);
}
vector<int> preorderTraversal(TreeNode* root) {
//递归版本
vector<int> v;
search(root,v);
return v;
}
};
/**
* Definition for a binary tree node.
* 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) {
//迭代版本
vector<int> v;
stack<TreeNode*> s;
s.push(root);
while(!s.empty()){
TreeNode *node = s.top();
s.pop();
if(!node) continue;
v.push_back(node->val);
if(node->right) s.push(node->right);
if(node->left) s.push(node->left);
}
return v;
}
};
/**
* Definition for a binary tree node.
* 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) {
//大佬提出的只压右节点,优化了迭代,大佬们牛批
vector<int> v;
TreeNode *node = root;
stack<TreeNode*> s;
while(node || !s.empty()){
if(!node){
node = s.top();
s.pop();
}
if(node->right) s.push(node->right);
v.push_back(node->val);
node = node->left;
}
return v;
}
};