vector<int> preorderTraversal(TreeNode *root) { stack<TreeNode*>StackTree; vector<int>data;if(root==NULL)returndata; StackTree.push(root);while(!StackTree.empty()) { TreeNode*CurrentNode=StackTree.top(); StackTree.pop(); data.push_back(CurrentNode->val);if(CurrentNode->right!=NULL) ...
Recursive solution ---O(n)time andO(n)space (considering the spaces of call stack); Morris traversal ---O(n)time andO(1)space!!! Iterative solution using stack: 1vector<int> preorderTraversal(TreeNode*root) {2vector<int>nodes;3stack<TreeNode*>toVisit;4TreeNode* curNode =root;5while...
classSolution {public: vector<int> preorderTraversal(TreeNode *root) { vector<int>res; stack<TreeNode *>s;if(root == NULL){//空树returnres; } s.push(root);//放树根while(!s.empty()){ TreeNode*cc =(); s.pop(); res.push_back(cc->val);//访问根if(cc->right !=NULL){ s.p...
第一种思路就是利用栈去模拟上边的递归。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Stack<TreeNode>stack=newStack<>();TreeNodecur=root;while(cur!=null||!stack.isEmpty()){if(cur!=null){list.add(cur.val);stack.push(cur);cur=cur.left;//考虑左子...
Given a binary tree, return the preorder traversal of its nodes’ values. For example: Given binary tree [1,null,2,3], 1 \ 2 / 3 1. 2. 3. 4. 5. return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively?
INORDER AND PREORDER TRAVERSAL and DISPLAY OF EXISTING BINARY TREE IN COMPUTER MEMORYP K KumaresanJournal of Harmonized Research
* public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<Integer>();if(root==null){returnret;}Stack<TreeNode>st=newStack<TreeNo...
[Int]valstack=Stack[TreeNode]()varcurr=rootwhile(stack.nonEmpty||curr!=null){while(curr!=null){stack.push(curr)rs.append(curr.value)curr=curr.left}valnode=stack.pop()curr=node.right}rs.toList}defpreorderTraversalV4(root:TreeNode):List[Int]={if(root==null)returnNilvalrs=ListBuffer.empty...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
Given sequence represents preorder traversal of a BST The time complexity of the above solution is O(n), where n is the size of the BST, and requires space proportional to the tree’s height for the call stack. Also See: Build a Binary Search Tree from a preorder sequence Build a ...