#LeetCode# #C++# Binary Tree Preorder Traversal

给一个二叉树，返回前驱遍历的节点值

Note：递归容易，迭代呢？

方法一：递归

1.存root->val

2.递归存root->left

3.递归存root->right

Status: Accepted

Runtime: 0 ms

class Solution {
public:
void fillresult(TreeNode* root, vector<int> &result){
if(root == NULL) return;
result.push_back(root->val);
fillresult(root->left, result);
fillresult(root->right, result);
}

vector<int> preorderTraversal(TreeNode* root) {
if(root == NULL) return vector<int>();
vector<int> result;
fillresult(root, result);
return result;
}
};

方法二：迭代

1.stack存入root，作为开头

2.stack不为空，循环3、4

3.top取一个点ptop，存入vector

4.pop掉ptop，stack存入非空的ptop->right，stack存入非空的ptop->left

5.return vector

Status: Accepted

Runtime: 0 ms

class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> result;
if(root == NULL) return result;

stack<TreeNode*> stk;
TreeNode* pNode = NULL;
stk.push(root);
while(!stk.empty()){
pNode = stk.top();
result.push_back(pNode->val);
stk.pop();
if(pNode->right){
stk.push(pNode->right);
}
if(pNode->left)
stk.push(pNode->left);
}
return result;

}
};