LeetCode | 144. Binary Tree Preorder Traversal

 

题目:

Given the root of a binary tree, return the preorder traversal of its nodes' values.

 

Example 1:

LeetCode | 144. Binary Tree Preorder Traversal

Input: root = [1,null,2,3]
Output: [1,2,3]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

LeetCode | 144. Binary Tree Preorder Traversal

Input: root = [1,2]
Output: [1,2]

Example 5:

LeetCode | 144. Binary Tree Preorder Traversal

Input: root = [1,null,2]
Output: [1,2]

 

Constraints:

The number of nodes in the tree is in the range [0, 100].-100 <= Node.val <= 100

 

Follow up:

Recursive solution is trivial, could you do it iteratively?

 

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 backPreorder(TreeNode* root, vector<int>& res) {
if(root == NULL)
return;
res.push_back(root->val);
backPreorder(root->left, res);
backPreorder(root->right, res);
return;
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
backPreorder(root, res);
return res;
}
};

 

 

 

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LeetCode | 144. Binary Tree Preorder Traversal

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