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…

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…

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…

Binary Tree 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5573 Description The Old Frog King lives on the root of an infinite tree. According to the law, each node should connect to exactly two nodes on the next level, forming a full binary tree. S…

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…

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…

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…

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 / \ 9 20…

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 / \ 9 2…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:  You may assume that duplicates do not exist in the tree. 利用前序和中序遍历构造二叉树的思想与利用中序和后续遍历的思想一样,不同的是,全树的根节点为preorder数组的第一个元素,具体分析过程参照利用中序和后续构造二叉树.这两道题是从二叉树的前序遍历.中序遍历.后续遍历这三种遍…