144. Binary Tree Preorder Traversal

Posted 2021-11-29 dmndxld

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

Example:

Input: [1,null,2,3]
   1
         2
    /
   3

Output: [1,2,3]

My idea:ricursion

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        res = []
        res.append(root.val)
        if root.left:
            left = self.preorderTraversal(root.left)
            res.extend(left)
        if root.right:
            right = self.preorderTraversal(root.right)
            res.extend(right)
        return res

 

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        res = []
        res.append(root.val)
        res.extend(self.preorderTraversal(root.left))
        res.extend(self.preorderTraversal(root.right))
        return res

 

 

OTHERS:iretator

/**
 * 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) {
        stack<TreeNode*> s;
        vector<int> ans;
        while(true){
            while(root){
                ans.push_back(root->val);
                if(root->right) s.push(root->right);
                root = root->left;
            }
            if(s.empty()) break;
            root = s.top();
            s.pop();
        }
        return ans;
    }
};