Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is...
Given a binary tree, return theinordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what"{1,#,2,3}"means?> read more on how binary tree is seriali...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> inorderTraversal(TreeNode *root) { vector<int>res; inorderTrav(root,res);returnres; }voidinorderTrav(TreeNode *root,vector<int> &res) {if(root==NULL)return; inorderTrav(root->le...
vector<int> inorderTraversal(TreeNode* root) { vector<int> res; inorder(res,root); return res; } private: void inorder(vector<int>& res,TreeNode* root){ if(root==NULL) return; inorder(res,root->left); res.push_back(root->val); inorder(res,root->right); } }; 1. 2. 3. ...
publicList<Integer>inorderTraversal3(TreeNoderoot){List<Integer>ans=newArrayList<>();TreeNodecur=root;while(cur!=null){//情况 1if(cur.left==null){ans.add(cur.val);cur=cur.right;}else{//找左子树最右边的节点TreeNodepre=cur.left;while(pre.right!=null&&pre.right!=cur){pre=pre.right;}...
Given a binary tree, return theinordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[1,3,2]. 1 /** 2 * Definition for binary tree 3 * struct TreeNode { ...
Binary Trees Preorder Traversal of a tree both using recursion and Iteration <-> Binary Trees Postorder Traversal of a tree both using recursion and Iteration <-> Binary Trees Left View of a tree <-> Binary Trees Right View of Tree https://leetcode.com/problems/binary-tree-right-side-view...
you'll often need some way to get access to each object in the collection. In fact, depending on the collection, you may want several ways to access each object such as front to back, back to front, preorder or postorder. To keep the collection simple, the traversal code itself is oft...
you'll often need some way to get access to each object in the collection. In fact, depending on the collection, you may want several ways to access each object such as front to back, back to front, preorder or postorder. To keep the collection simple, the traversal code itself is oft...
Java Example public class Main { public static void main(String[] args) { int[] myarray = {12,4,3,1,15,45,33,21,10,2}; System.out.println("Input list of elements ..."); for(int i=0;i<10;i++) { System.out.print(myarray[i] + " "); ...