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 the inorder traversal 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?
(referrence:ProgramCreek) The key to solve inorder traversal of binary tree includes the following: The order of "inorder" is: left child -> parent -> right child Use a stack to track nodes Understand when to push node into the stack and when to pop node out of the stack Note that ...
2.2 last.right 不为 null,说明之前已经访问过,第二次来到这里,表明当前子树遍历完成,保存 cur 的值,更新 cur = cur.right public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left ...
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 { ...
树的中序遍历。先不断压入左结点至末尾,再访问,再压入右结点。注意和先序遍历的比较 vector<int> inorderTraversal(TreeNode *root) { vector<int>result; stack<TreeNode *>s; TreeNode*p =root;while(!s.empty() || p !=nullptr) {if(p !=nullptr) ...
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...
Subtree of Another Tree -https://leetcode.com/problems/subtree-of-another-tree/ Construct Binary Tree from Preorder and Inorder Traversal -https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ Validate Binary Search Tree -https://leetcode.com/problems/validate...
*/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.val);current=current.right;}else{// has left, then find the rightmost of left su...
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...