int*inorderTraversal(structTreeNode* root,int* returnSize){ *returnSize =0;int* res =malloc(sizeof(int) *501);structTreeNode**stk=malloc(sizeof(structTreeNode*) *501);inttop =0;while(root !=NULL|| top >0) {while(root !=NULL) { stk[top++] = root; root = root->left; } root...
binary-tree-inorder-traversal /** * * @author gentleKay * 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?
publicList<Integer>inorderTraversal(TreeNoderoot){List<Integer>ans=newArrayList<>();Stack<TreeNode>stack=newStack<>();TreeNodecur=root;while(cur!=null||!stack.isEmpty()){//节点不为空一直压栈while(cur!=null){stack.push(cur);cur=cur.left;//考虑左子树}//节点为空,就出栈cur=stack.pop()...
Given abinary tree, return theinordertraversal of its nodes' values. 知识点: Recursion Iteration Threaded binary tree 二刷: 要求的iteration: 最初做法: class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: if not root: return [] ans = [] stack = [root] l = [0]...
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> inorderTraversal(TreeNode *root) { 13 vector<int> res; 14 if(root==NULL) return res; 15 stack<TreeNode*> s; ...
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...
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:javascript 代码运行次数:0 运行 AI代码解释 ...
* 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);current=current.right...
** Inorder Traversal: left -> root -> right ** Preoder Traversal: root -> left -> right ** Postoder Traveral: left -> right -> root 记忆方式:order的名字指的是root在什么位置。left,right的相对位置是固定的。 图片来源:https://leetcode.com/articles/binary-tree-right-side-view/ ...
Unfortunately, the .NET Framework does not contain a binary tree class, so in order to better understand binary trees, let's take a moment to create our own binary tree class. The First Step: Creating a Base Node Class The first step in designing our binary tree class is to create a ...