【LeetCode】144. Binary Tree Preorder Traversal (3 solutions)

Binary Tree Preorder Traversal

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?

解法一：递归

/**
* 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) {
vector<int> ret;
Helper(ret, root);
return ret;
}
void Helper(vector<int>& ret, TreeNode* root)
{
if(root)
{
ret.push_back(root->val);
Helper(ret, root->left);
Helper(ret, root->right);
}
}
};

解法二：借助栈的非递归，需要记录每个节点是否访问过

/**
* 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> ret;
if(root == NULL)
return ret;
stack<TreeNode*> stk;
unordered_map<TreeNode*, bool> visited;
stk.push(root);
visited[root] = true;
while(!stk.empty())
{
TreeNode* top = stk.top();
stk.pop();
ret.push_back(top->val);
if(top->right != NULL && visited[top->right] == false)
{
stk.push(top->right);
visited[top->right] = true;
}
if(top->left != NULL && visited[top->left] == false)
{
stk.push(top->left);
visited[top->left] = true;
}
}
return ret;
}
};

解法三：借助栈的非递归，无需记录每个节点是否访问过。

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