classSolution { public List<Integer>inorderTraversal(TreeNode root) { List<Integer> traversal = new ArrayList<>(); forInOrderTraversal(root,traversal); returntraversal; } public static void forInOrderTraversal(TreeNode T,List<Integer>traversal){ if(T==null) return; forInOrderTraversal(T.left,...
preorder traversal: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> inorderTraversal(TreeNode *root) { // Start typi...
https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 给定两个整数数组preorder和inorder,其中preorder是二叉树的先序遍历,inorder是同一棵树的中序遍历,请构造二叉树并返回其根节点。 示例1: 输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] 输出...
def inorderTraversal(self, root): res, stack = [], [] while stack or root: if root: stack.append(root) root=root.left else: node = stack.pop() res.append(node.val) root = node.right return res 3. 后序遍历 145. 二叉树的后序遍历 后序遍历,输出为[1,2,3,4,5] 递归 def in...
*/publicList<Integer>inorderTraversal(TreeNode root){List<Integer>nodes=newArrayList<>(16);if(root==null){returnnodes;}//使用链表作为栈Deque<TreeNode>stack=newLinkedList<TreeNode>();while(root!=null||!stack.isEmpty()){//遍历左子树while(root!=null){stack.push(root);root=root.left;}//取...
preorderTraversal 函数: 这是主函数,用于执行前序遍历并返回结果数组。它首先调用 TreeSize 函数(虽然这里没有给出 TreeSize 的实现,但我们可以假设它的功能是计算树的节点数)来计算树的节点数,然后动态分配一个足够大的整数数组来存储结果。接着,它调用 _prevOrder 函数来执行前序遍历,并填充数组。最后,它设置...
inorderTraversal(root.right); } return res; } } 时间复杂度:O(n),其中n为二叉树节点的个数。二叉树的遍历中每个节点会被访问一次且只会被访问一次。 空间复杂度:O(n)。空间复杂度取决于递归的栈深度,而栈深度在二叉树为一条链的情况下会达到O(n)的级别。
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"Object is currently in use elsewhere" error for picturebox "Parameter is not valid" - new Bitmap() "Recursive write lock acquisitions not allowed in this mode.? "Settings" in DLL project properties and app.config file "The function evaluation requires all threads to run" while accessing m...
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