1.题目描述 2.分析 找出最大元素,然后分割数组调用. 3.代码 TreeNode* constructMaximumBinaryTree(vector<int>& nums) { int size = nums.size(); ) return NULL; TreeNode *dummy = ); maxcont(dummy->left, , size-, nums); return dummy->left; } void maxcont(TreeNode* &…

Maximum Binary Tree 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/maximum-binary-tree/description/ Description Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in…

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number…

题目: Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum nu…

题目: 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? 说明:1)下面有两种实现:递归(Recursive )与非递归(迭代iterati…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to the…

题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 说明: 1)二叉树可空 2)思路:a.根据前序遍历的特点, 知前序序列(PreSequence)的首个元素(PreSequence[0])为二叉树的根(root),  然后在中序序列(InSequence)中查找此根(…

Given an integer array with no duplicates. A maximum tree building on this array is defined as follow: The root is the maximum number in the array. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number…

1.题目描述 2.问题分析 使用层序遍历 3.代码 vector<int> v; vector<int> rightSideView(TreeNode* root) { if (root == NULL) return v; queue<TreeNode*> q; q.push(root); while (!q.empty()) { int size = q.size(); ; i < size; i++) { TreeNode *node = q.front()…

1.题目描述 2.问题分析 使用递归 3.代码 TreeNode* pruneTree(TreeNode* root) { if (root == NULL) return NULL; prun(root); return root; } void prun(TreeNode *root) { if (root == NULL) return; if (!has_ones(root->left)) root->left = NULL; if (!has_ones(root->right)…