# LeetCode – Inorder Successor in BST (Java)

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

Java Solution 1

```public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { Stack<TreeNode> stack = new Stack<TreeNode>(); if(root==null || p==null) return null;   stack.push(root); boolean isNext = false; while(!stack.isEmpty()){ TreeNode top = stack.pop();   if(top.right==null&&top.left==null){ if(isNext){ return top; }   if(p.val==top.val){ isNext = true; } continue; }   if(top.right!=null){ stack.push(top.right); top.right=null; }   stack.push(top);   if(top.left!=null){ stack.push(top.left); top.left=null; } }   return null; }```

Time is O(n), Space is O(n).

Java Solution 2

```public TreeNode inorderSuccessor(TreeNode root, TreeNode p) { if(root==null) return null;   TreeNode next = null; TreeNode c = root; while(c!=null && c.val!=p.val){ if(c.val > p.val){ next = c; c = c.left; }else{ c= c.right; } }   if(c==null) return null;   if(c.right==null) return next;   c = c.right; while(c.left!=null) c = c.left;   return c; }```

Time is O(log(n)) and space is O(1).

Category >> Algorithms
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• Udaydeep Thota

Simple and easy to understand:

// Just do pre order traversal and while processing the node, if it is find node :
// case 1: if node has right link, go to the left most of the right link and return the data
// case 2: else pop the stack and return it (which means the next highest value of the node)
// Time complexity : O(n)
public static BinaryTreeNode findInOrderSuccessor(BinaryTreeNode root, BinaryTreeNode findNode) {

if(root==null || findNode==null)
return null;

Stack nodeStoreStack = new Stack();
BinaryTreeNode current = root;

while (!nodeStoreStack.isEmpty() || current!=null) {

if(current!=null) {
}

else {

BinaryTreeNode temp = nodeStoreStack.pop();

if(temp==findNode) {

}
return successor;

}
else if(!nodeStoreStack.isEmpty()) {
return nodeStoreStack.pop();
}

}

}

}

return null;

}