在递归调用方法的情况下返回

Question

I have a piece of code that I need insight on, I don't want to know the correct solution of the problem. I just want to know why is my concept failing. So the function is to check if a binary tree is BST or not. I need to return 1 in case it is and 0 otherwise. My code is as below

int isBST(Node root) {
 if(root == null)
  return 1;
 if(root.left!=null) {
  if(root.left.data<root.data)
    isBST(root.left);
  else
   return 0; // line:a
 }
 if(root.right!=null) {
  if(root.right.data<root.data)
    isBST(root.right);
  else
   return 0;
 }
return 1;
}

For this piece, when i have a binary tree as follows:

5
 \
  7
 /
8

I expect it to reach 8 value and break at line:a but it return me 1 instead of 0. Now I know 0 gets returned to the parent calling method. But is it not terminating because I have made isBST call without capturing the return value? Please dont point out if there are anyother bugs.

Answer 1

For the general case, your approach won't work. The way to test if a tree is a BST is to recursively check if the current node is greater than the maximum element of the left subtree, and smaller than the minimum element of the right subtree. Also, you're missing the returns in the recursive calls:

return isBST(root.left);
...
return isBST(root.right);

By the way, why are you returning 0 or 1 for this? use false or true instead, changing the return type to boolean .

Answer 2

您应该检查正确的数据是否大于当前数据，并返回递归调用的值

if(root.right.data>root.data)

Answer 3

I see the following issues:

• If (and only if) isBST(root.left) or isBST(root.right) is false, you need to return 0 (btw, why are you not using booleans?) immediately.
• The condition root.right.data<root.data should be inverted: root.right.data>=root.data .

So here's the modified code (keeping the int return type):

int isBST(Node root) {
 if(root == null)
  return 1;
 if(root.left!=null) {
  if(root.left.data<root.data) {
    if (isBST(root.left) == 0)
      return 0;
  } else
   return 0;
 }
 if(root.right!=null) {
  if(root.right.data>=root.data) {
    if (isBST(root.right) == 0) {
      return 0;
    }
  } else
   return 0;
 }
return 1;
}