膜这场比赛的 \(rk1\) \(\color{black}A\color{red}{lex\_Wei}\) 这题应该是这场比赛最难的题了 容易发现,二叉树的下一层不会超过这一层的 \(2\) 倍,所以我们先构造出来一颗尽量满的二叉树,然后慢慢向下调整,调整的方法是从最上面一个一个弄下来. 然后你慢慢调整的复杂度最多是 \(d\) ,复杂度 \(O(d)\) #include <bits/stdc++.h> using namespace std ; const int maxn = 5e3…

Solution 预处理出 \(i\) 个点组成的二叉树的最大答案和最小答案 递归做,由于只需要构造一种方案,我们让左子树大小能小就小,因此每次从小到大枚举左子树的点数并检验,如果检验通过就选定之 现在还需要确定左右子树各分配多少答案上去,一种构造性的想法是,要么让左子树选最小,要么让右子树选最大.对于任意一种其它方案,我们可以通过把左子树上的答案不断移到右子树上,直到某一边达到界限,故等价. 打错变量名查半天-- #include <bits/stdc++.h> using namespac…

ACM思维题训练集合 You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d. A tree is a connected graph without cycles. A rooted tre…

function createNode(value) { return { value, left: null, right: null }; } function BinaryTree(val) { return { root: null, nodes: [], add(val) { const node = createNode(val); if (!this.root) { this.root = node; } else { this.downShift(node); } this.no…

http://www.geeksforgeeks.org/full-and-complete-binary-tree-from-given-preorder-and-postorder-traversals/ #include <iostream> #include <vector> #include <algorithm> #include <queue> #include <stack> #include <string> #in…

题意:给定节点数n和所有节点的深度总和d,问能否构造出这样的二叉树.能,则输出“YES”,并且输出n-1个节点的父节点(节点1为根节点). 题解:n个节点构成的二叉树中,完全(满)二叉树的深度总和最小,单链树(左/右偏数)的深度总和最大.若d在这个范围内,则一定能构造出来:否则一定构造不出来. 1.初始构造一颗单链树,依次把底部的节点放入上面的层,直到满足深度总和为d 2.若当前深度总和sum > d,则先拿掉底端节点. 拿掉后,若sum依然比d大,就直接把底端节点放入有空位的最上层: 拿掉后s…

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题要求从中序和后序遍历的结果来重建原二叉树,我们知道中序的遍历顺序是左-根-右,后序的顺序是左-右-根,对于这种树的重建一般都是采用递归来做,可参见我之前的一篇博客Convert Sorted Array to Bin…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 这道题要求用先序和中序遍历来建立二叉树,跟之前那道Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树原理基本相同,针对这道题,由于先…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. class Solution { public: TreeNode *buildTree(vector<int>& inorder, int in_left,int in_right, vector<int&…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. Subscribe to see which companies asked this question /** * Definition for a binary tree node. * struct TreeNode…