benbenab的专栏

Given preorder and inorder 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

Unique Binary Search Trees Given n, how many structurally unique BST's (binary search trees) that store values 1...n?class Solution {public:    int numTrees(int n) {        // Star

1. 链表-->BST (借鉴smile的)/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; *//** * Definition

Given a binary tree, find the maximum path sum.The path may start and end at any node in the tree.For example:Given the below binary tree, 1 / \ 2 3Return 6.

search the most right node from root. If the most right root has no child, return its parent; otherwise return the largest number of its left child tree

void convert(TreeNode* root,TreeNode** preNode,TreeNode** head){ if (root==NULL) return; convert(root->left,preNode,head); if (*preNode!= NULL){ root->left = *preNode; (*preNode)->right = ro

要求路径可以从任何位置开始void checkPath(int sum,vector result,vector>& results ){ vector resultTemp; for(int i = result.size()-1;i>=0;i--){ sum -= result.at(i)->val; resultTemp.insert(resultTemp.begin()

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).For example, this binary tree is symmetric: 1 / \ 2 2 / \ / \3 4 4 3But the f

假设一棵树的每个node都有不同的标号，给定后序遍历和前序遍历，不同的树有多少种？思路：给定中序遍历能确定树的具体结构，是因为可以确定到底是左子树还是右子树，而之给定preorder和pastorder，则不能确定，这也是为什么会出现不同树结构的原因。只有 post 和 pre 那么主要问题就是没有办法处理只有一个subtree的 node，因为 这种情况 不知道subtree 究竟是这个

Given a linked list, determine if it has a cycle in it.Follow up:Can you solve it without using extra space?