Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 见剑指offer重建二叉树 class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) {
Problem： Given a binary tree, return the inorder traversal of its nodes' values. Explanation： 中序遍历给定二叉树 My Thinking： 使用常规递归进行中序遍历，不过这里需要添加一个函数，才能将结果列表作为参数传递。 My Solution： class Solution { public List<Integer> in
1、题目描述 2/问题分析 利用中序遍历，然后重新构造树。 3、代码 1 TreeNode* increasingBST(TreeNode* root) { 2 if (root == NULL) 3 return NULL; 4 vector<int> v; 5 inorder(root,v); 6 7 TreeNode* dummy
题目： https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/ 题意： 给定一个二叉树，返回它的中序 遍历。 思路： 经典题目了，递归版和非递归版，时间复杂度均为O(n)O(n)O(n) 代码： 递归版： class Solution { public: vector<int> inorderTraversal(TreeNode* root
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a node contains only nodes with key
class Solution { public: vector<int> modes; int maxCnt = 0; int curCnt = 0; int curNum = 0; vector<int> findMode(TreeNode* root) { if (!root) { return modes; } curNum = root->val;
73. Construct Binary Tree from Preorder and Inorder Traversal/ 本题难度: Medium Topic: Binary Tree Description Given preorder and inorder traversal of a tree, construct the binary tree. Example Example 1: Input: [], [] Output: null Example 2: Input: in-orde
算法描述： Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 /
算法描述： Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3
