*/// 由于主函数的形式已经符合分治法需要的形式(具有合适的返回值),直接使用主函数做为递归函数vector<int>preorderTraversal(TreeNode * root){//递归三要素之定义// write your code herevector<int> result;if(root == nullptr) {returnresult;// 递归三要素之出口}// 获取左子树的遍历结果vector<int> ...
classSolution {public: vector<int> preorderTraversal(TreeNode*root) {if(!root)return{}; vector<int>res; stack<TreeNode*>s{{root}};while(!s.empty()) { TreeNode*t =s.top(); s.pop(); res.push_back(t->val);if(t->right) s.push(t->right);if(t->left) s.push(t->left); ...
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...
class Solution(object): def _preorderTraversal(self, root, result): if root: result.append(root.val) self._preorderTraversal(root.left, result) self._preorderTraversal(root.right, result) def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ if root ==...
解法三 Morris Traversal 上边的两种解法,空间复杂度都是O(n),利用 Morris Traversal 可以使得空间复杂度变为O(1)。 它的主要思想就是利用叶子节点的左右子树是null,所以我们可以利用这个空间去存我们需要的节点,详细的可以参考94 题中序遍历。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list...
INORDER AND PREORDER TRAVERSAL and DISPLAY OF EXISTING BINARY TREE IN COMPUTER MEMORYP K KumaresanJournal of Harmonized Research
Morris traversal: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<...
[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[Int]valstack=Stack...
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...
每天一算:Binary Tree Preorder Traversal 程序员吴师兄 + 关注 预计阅读时间2分钟 6 年前 LeetCode上第144 号问题:二叉树的前序遍历 题目 给定一个二叉树,返回它的 前序 遍历。 示例: 输入: [1,null,2,3] 输出: [1,2,3] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题思路 用栈(Stack)...