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...
root.left =newTreeNode(2); root.left.left =newTreeNode(1); root.left.right =newTreeNode(3); root.right =newTreeNode(6); root.right.left =newTreeNode(5); root.right.right =newTreeNode(7); root.right.right.right =newTreeNode(8); System.out.println(Main21.inorderTraversal(root))...
*@return: Inorder in ArrayList which contains node values.*/publicArrayList<Integer>inorderTraversal(TreeNode root) {//write your code hereArrayList<Integer> rst =newArrayList<Integer>();if(root ==null) {returnrst; } traversal(rst, root);returnrst; }privatevoidtraversal(ArrayList<Integer>rst, ...
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;}...
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); ...
Leetcode 94.二叉树的中序遍历(Binary Tree Inorder Traversal) Leetcode 94.二叉树的中序遍历 1 题目描述(Leetcode题目链接) 给定一个二叉树,返回它的中序遍历。 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 2 题解 中序遍历就是对每个节点来说,先访问节点的左孩子,再访问本节点,再...
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 { ...
每天一算:Binary Tree Inorder Traversal 程序员吴师兄 + 关注 预计阅读时间2分钟 6 年前 LeetCode上第94 号问题:二叉树的中序遍历 题目 给定一个二叉树,返回它的 中序 遍历。 示例: 输入: [1,null,2,3] 输出: [1,3,2] 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)的...
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...
* TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>preorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){if(current.left==null){result.add(current....