1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12vector<int> preorderTraversal(TreeNode*root) {13vector<int>rVec;14pr...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> preorderTraversal(TreeNode*root) { vector<int>result;if(root ==nullptr...
public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<>(); preorderTraversalHelper(root, list); return list; } private void preorderTraversalHelper(TreeNode root, List<Integer> list) { if (root == null) { return; } list.add(root.val); preorderTrav...
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...
二叉树的前序遍历(Binary Tree Preorder Traversal) lintcode:题号——66,难度——easy 2 描述 给出一棵二叉树,返回其节点值的前序遍历。 名词: 遍历 按照一定的顺序对树中所有节点进行访问的过程叫做树的遍历。 前序遍历 在二叉树中,首先访问根结点然后遍历左子树,最后遍历右子树,在遍历左右子...
val), printPreorder(root.left) printPreorder(root.right) # Driver code root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(5) print("Preorder traversal of binary tree is") printPreorder(root) print("\nInorder traversal of ...
15、课程:树(下).10、练习—Iterative Postorder Traversal -- -- 10:33 App 15、课程:树(下).7、练习—Iterative Get和Iterative Add 2 -- 12:36 App 15、课程:树(下).13、练习—Construct Binary Tree from Preorder and Inorder Traversal 1 -- 9:07 App 15、课程:树(下).3、练习—Floor and...
* TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);...
node=stack.pop()curr=node.rightreturnrsdefpreorderTraversalV4(root):ifnotroot:return[]rs=[]stack=[(root,0)]whilestack:node,visited=stack.pop()ifnode:ifvisited:rs.append(node.val)else:stack.append((node.right,0))stack.append((node.left,0))stack.append((node,1))returnrsdefpreorderTraver...
The pre-order and post-order traversal of a Binary Tree is A 、B 、C and C 、B 、A, respectively. The number of trees that satisfy the conditions is . A.1 B.2 C.3 D.4 暂无答案