记录点滴

题目：Given a binary tree, find the maximum path sum.For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connecti

题目：Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.代码：/** * Definition for undirected graph. * struct UndirectedGraphNode { * int label;

观察：            A         /      \       B      C      /   \   D    Epreorder: A B D E Cinorder:   D B E A C  规律：对于inorder序列，最左边的一定是整个树的leftmost。这样，从左☞右遍历pre，直到pre[i] == in[0], 便找到了le

题目：Given a binary tree, determine if it is height-balanced.For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node neve

题目：Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.For example:Given the below binary tree and sum = 22, 5

题目：Given a binary tree, flatten it to a linked list in-place.For example,Given 1 / \ 2 5 / \ \ 3 4 6The flattened tree should look like:

题目：Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }Populate each next pointer to point to its next rig

题目：Follow up for problem "Populating Next Right Pointers in Each Node".What if the given tree could be any binary tree? Would your previous solution still work?Note:You may only use

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.The brackets must close in the correct order, "()" and "()[]{}" are all va

方法一：递归/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is

Given a collection of intervals, merge all overlapping intervals.For example,Given [1,3],[2,6],[8,10],[15,18],return [1,6],[8,10],[15,18]./** * Definition for an interval. * struct I

平衡查找树在一颗含有N个结点的树中，我们希望树高为~lgN，这样我们就能保证所有查找都能在~lgN此比较内结束，就和二分查找一样。不幸的是，在动态插入中保证树的完美平衡的代价太高了。我们放松对完美平衡的要求，使符号表API中所有操作均能够在对数时间内完成。 2-3查找树为了保证查找树的平衡性，我们需要一些灵活性，因此在这里我们允许树中的一个结点保

1.结论运行效率方面：unordered_map最高，hash_map其次，而map效率最低单提供了有序的序列。占用内存方面：hash_map内存占用最低，unordered_map其次(数量少时优于hash_map），而map占用最高。需要无序容器时候用unordered_map，有序容器时候用map。2.原理map的内部实现是二叉平衡

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */clas

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */clas

题目：Given a binary tree, return the inorder traversal of its nodes' values.For example:Given binary tree [1,null,2,3], 1 \ 2 / 3return [1,3,2].解法：Hi,

题目复制带有随机指针的链表：一个单链表除了next指针外还有一个random指针随机指向任何一个元素（可能为空） 《剑指offer》上的面试题26分析方法一： map，先按普通方法复制链表，再两个链表同时走复制random（旧节点a，新节点a’）a'->random=map[a->random](空指针单独处理) 方法二： 插入：每个节点后面插入一个自身的“复本

Given a binary tree, determine if it is a valid binary search tree (BST).class Solution {public: bool isValidBST(TreeNode* root) { TreeNode* prev = NULL; return validate(r

一、线索二叉树的原理    通过考察各种二叉链表，不管儿叉树的形态如何，空链域的个数总是多过非空链域的个数。准确的说，n各结点的二叉链表共有2n个链域，非空链域为n-1个，但其中的空链域却有n+1个。如下图所示。    因此，提出了一种方法，利用原来的空链域存放指针，指向树中其他结点。这种指针称为线索。    记ptr指向二叉链表中的一个结点，以下是建立线索的规

题目：基础题，代码里给出了三种解答方法，其中Morris法是O(1)space。Two elements of a binary search tree (BST) are swapped by mistake.Recover the tree without changing its structure.Note:A solution using O(n) spa

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

题目：Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).For example:Given binary tree [3,9,20,null,null,15,7], 3 /

vectorvectorint> > zigzagLevelOrder(TreeNode* root) { if (root == NULL) { return vectorvectorint> > (); } vectorvectorint> > result; queue nodesQueue; nodesQueue.push(root

1，my solution/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {}

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).You may assume that the intervals were initially sorted according to their start times.E