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/*Key to solve: BFS Trough Tree level by level(BFS) 1.treeCurr List: store TreeNode of current level 2.curr List: store the value of TreeNode of current level into curr List 3.Use a I

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?For example,Given n = 3, there are a total of 5 unique BST's. 1 3 3 2 1 \

Given n, generate all structurally unique BST's (binary search trees) that store values 1...n.For example,Given n = 3, your program should return all 5 unique BST's shown below. 1 3

import java.util.Stack;import java.util.List;import java.util.ArrayList;//inorder : left -> middle -> right//using stack to track nodes//Time Complexity: O(n)//Space Complexity: O(n)public clas

Key to solve: BFS by ArrayList    Create a TreeNode list that including all the nodes for each level, terminate loop if list is empty    Build another Integer list that including value of each lev

Normal Approach: In_Order Approach    Idea: simply do In_Order traversal, then check    If In_Order traversal of a binary tree is sorted,    then the binary tree is BST. else notpublic static

/* Key to solve: DFS, Recursion, BackTracking similar with previous problem of Path Sum, instead of print out all path that meet equal to user inout "sum" 1.Run DFS 2.Backtracking

/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ /* Key to solve: recursion,

/*Given a binary tree, return the postorder traversal of its nodes' values.For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3return [3,2,1].Note: Recursive solution is trivia

/*Given a binary tree, return the preorder traversal of its nodes' values.For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3return [1,2,3].Note: Recursive solution is trivial

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:Only one letter can be changed at a timeEach intermediate word m

/* Key to Solve: * Recursion, * array as a reference object to store max sum The max path sum is base on 4 cases possibility: 1. root-self. 2. L-subSum + R-subSum + root 3. L-su

/* Key to solve: DFS 1.find length of LinkedList 2. Need to set the mid point of Linkedlist as root in order to construct Binary Search Tree 3.recursively construct left sub tree from

Given preorder and inorder traversal of a tree, construct the binary tree.Note:You may assume that duplicates do not exist in the tree.

/* Solution1: key to solve: DFS, recursion 1. */ public static int minDepthDFS(TreeNode root) { //base case if (root == null) return 0; //recursion

Given inorder and postorder traversal of a tree, construct the binary tree.Note:You may assume that duplicates do not exist in the tree.