前序遍历: 根->左->右 递归方法 publicvoidpreorder(TreeNode root,List<Integer> res){if(root ==null){return; } res.add(root.val); preorder(root.left,res); preorder(root.right,res);return; }publicList<Integer>preorderTraversal(TreeNode root){ List<Integer> res =newArrayList<Integer>();...
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...
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...
Given a binary tree, return thepreordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[1,2,3]. Note:Recursive solution is trivial, could you do it iteratively?
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<...
public:vector<int>preorderTraversal(TreeNode*root){vector<int>result;stack<TreeNode*>st;if(root==NULL)returnresult;st.push(root);while(!st.empty()){TreeNode*top=st.top();st.pop();result.push_back(top->val);if(top->right)st.push(top->right);if(top->left)st.push(top->left);}...
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...
} void tree::preorder_traversal(node *r){ if(r != NULL){ //When root is present, visit left - root - right cout << r->value << " "; preorder_traversal(r->left); preorder_traversal(r->right); } } node *tree::insert_node(node *root, int key){ if(root == NULL){ retu...
Here, we are going to see given an array how can we check whether the given array can represent preorder traversal of a binary search tree or not. For example, Let's take two example Where one is a valid preorder traversal and another is not ...