1-10的中序二叉搜索树遍历

Question

I am doing a simple binary search tree implementation in C++. I found that it works for most test cases but I am confused with a test case where I create a tree and add 1,2,3,4,5,6,7,8,9,10 in that order. The inorder traversal comes out to be 1,10,2,3,4,5,6,7,8,9. My understanding is that the inorder traversal will print the elements in sorted order which would be 1,2,3,4,5,6,7,8,9,10. However, either this assumption is incorrect or my code is printing an incorrect output. Please let me know if my output is correct or incorrect, and why if it is correct. Thank you.

Answer 1

If you are not sure about how in order binary tree traversal works then look at the below tree and the explanation.

(NOTE: I have this done only for 5 number's).

 /* Constructed binary tree is 
              1 
            /   \ 
          2      3 
        /  \ 
      4     5 
    */

Step 1 Creates an empty stack: S = NULL

Step 2 sets current as address of root: current -> 1

Step 3 Pushes the current node and set current = current->left until current is NULL
     current -> 1
     push 1: Stack S -> 1
     current -> 2
     push 2: Stack S -> 2, 1
     current -> 4
     push 4: Stack S -> 4, 2, 1
     current = NULL

Step 4 pops from S
     a) Pop 4: Stack S -> 2, 1
     b) print "4"
     c) current = NULL /*right of 4 */ and go to step 3
Since current is NULL step 3 doesn't do anything. 

Step 4 pops again.
     a) Pop 2: Stack S -> 1
     b) print "2"
     c) current -> 5/*right of 2 */ and go to step 3

Step 3 pushes 5 to stack and makes current NULL
     Stack S -> 5, 1
     current = NULL

Step 4 pops from S
     a) Pop 5: Stack S -> 1
     b) print "5"
     c) current = NULL /*right of 5 */ and go to step 3
Since current is NULL step 3 doesn't do anything

Step 4 pops again.
     a) Pop 1: Stack S -> NULL
     b) print "1"
     c) current -> 3 /*right of 5 */  

Step 3 pushes 3 to stack and makes current NULL
     Stack S -> 3
     current = NULL

Step 4 pops from S
     a) Pop 3: Stack S -> NULL
     b) print "3"
     c) current = NULL /*right of 3 */  

Traversal is done now as stack S is empty and current is NULL.