Inorder successor in BST solution

Question

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

Note:if The given node has a no In-order successor in the tree, return null .

Solution--Iterative

Inorder result is a ascending array for BST. Thus, this problem can is transferred to find successor for a node in BST.

successor = first element is greater than target node.

Consider situations:

1. Target node has right child

2. Target node doesn ' t has right child

Trace back, find first node whose left child was in the path from target node to root.

For this situation:

1. If node has the parent pointer, we just use the parent pointer.

2. If node doesn ' t has the parent pointer, we use the stack.

1 /**2 * Definition for a binary tree node.3 * public class TreeNode {4 * int val;5 * TreeNode left;6 * TreeNode right;7 * TreeNode (int x) {val = x;}8  * }9  */Ten  Public classSolution { One      PublicTreeNode inordersuccessor (TreeNode root, TreeNode p) { A         //same question with finding successor in a BST -         //1. If P has right child, find minimum child in right subtree -         //2. If p doesn ' t has right child, find first ancestor whose left child is in the path from p to root. theTreeNode result =NULL; -         if(P.right! =NULL) { -result =P.right; -              while(Result.left! =NULL) { +result =Result.left; -             } +}Else { Adeque<treenode> stack =NewLinkedlist<treenode>(); atTreeNode current =Root; -             //push to Stack -              while(Current! = P && Current! =NULL) { - Stack.push (current); -                 if(P.val <current.val) { -Current =Current.left; in}Else { -Current =Current.right; to                 } +             } -             //Pop Stack theTreeNode prev =p; *              while(!Stack.isempty ()) { $TreeNode cur =Stack.pop ();Panax Notoginseng                 if(Cur.left = =prev) { -result =cur; the                      Break; +                 } APrev =cur; the             } +         } -         returnresult; $     } $}

Solution 2--Recursive

Find successor & predecessor in BST

Inorder successor in BST solution