Answers for "inorder binary tree"

0

inorder

void inorder(Node* root){
    if(root != NULL){
        inorder(root->left);
        cout<<root->data<<" ";
        inorder(root->right);
    }
}
Posted by: Guest on August-01-2021
0

inorder traversal

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        return dfs(root, list);
    }
    private List<Integer> dfs(TreeNode root, List<Integer> list)
    {
        if(root == null)
            return list;
        list = dfs(root.left, list);
        list.add(root.val);
        return dfs(root.right,list);
    }
}
Posted by: Guest on July-20-2021

Code answers related to "inorder binary tree"

Browse Popular Code Answers by Language