285. Inorder successor in BST

    /** 285. Inorder successor in BST * 2016-6-27 by Mingyang * Online Some code is too complex, my most simple and clear, is nothing more than a DFS, with a stack to do just fine, encountered after p * with a flag ma RK a bit, then you can return to the next inorder. * Note: Judging is!stack.empty () and not another method*/      PublicTreeNode inordersuccessor (TreeNode root, TreeNode p) {if(root==NULL)              return NULL; Stack<TreeNode> stack=NewStack<treenode>(); TreeNode s=Root; Booleanfindornot=false;  while(!stack.empty () | | s!=NULL){                if(s!=NULL) {Stack.push (s); S=S.left; }Else{TreeNode temp=Stack.pop (); if(findornot) {returntemp; }Else{                        if(temp==p) Findornot=true; } s=Temp.right; }            }            return NULL; }

285. Inorder successor in BST