03 2019 档案
摘要：Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path betwe
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摘要：Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two orzero sub-node. I
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摘要：Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might
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摘要：Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target. S
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摘要：Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subt
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摘要：Find the sum of all left leaves in a given binary tree. Solution1 :Recursive递归，注意：left leaves是左子叶，终端节点才是叶节点。所以只有9和15计算在内 class Solution { public int s
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摘要：Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array. class Solution { public List<Double> avera
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摘要：Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are n
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摘要：You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way. The null node needs to be rep
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摘要：* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }
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摘要：Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a
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摘要：Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. Assume a BST is define
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摘要：You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need t
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摘要：Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: Solution1: class Solution { public List<String> bina
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摘要：For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of
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摘要：Solution1: class Solution { public TreeNode invertTree(TreeNode root) { if (root == null) return null; TreeNode tmp = root.left; root.left = invertTre
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摘要：Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. F
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摘要：Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest l
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摘要：Given an array where elements are sorted in ascending order, convert it to a height balanced BST. class Solution { public TreeNode sortedArrayToBST(in
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摘要：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 dept
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摘要：Solution1:广度优先 class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { Queue<TreeNode> q = new LinkedList<>(); List<List<Integer
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摘要：Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest l
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摘要：Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). Solution1： class Solution { public boolean isSymmetric(
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摘要：Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical a
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摘要：Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes. Solution 1://先遍历左节点然后把左节点设为
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摘要：Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the roo
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摘要：Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only
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摘要：//Solution 1：递归 public List<Integer> inorderTraversal(TreeNode root) { List<Integer> list =new ArrayList<>(); addNode(list, root); return list;}privat
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摘要：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'
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摘要：Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n. For example,Given n = 3, your program should
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摘要：Solution 1: public List<List<Integer>> levelOrder(TreeNode root) { Queue<TreeNode> queue = new LinkedList<>(); List<List<Integer>> wrapList = new Link
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摘要：Solution 1： public class Solution { public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> sol = new ArrayList<>(); travel(r
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摘要：Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 见剑指offer重建二叉树
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摘要：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 Soluti
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摘要：given a binary tree, return the preorder traversal of its nodes' values. Solution 1://非递归 Solution 1://非递归 public class Solution { public List<Integer
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摘要：Given a binary tree, return the postorder traversal of its nodes' values. Solution 1: class Solution { public List<Integer> postorderTraversal(TreeNod
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摘要：Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. class Solution { public TreeNode sortedL
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摘要：//例如根节点为1，左2右3 class Solution { TreeNode prev = null; public void flatten(TreeNode root) {//先把最大的数设在root.right，然后剩下的数一个个往里加 if (root == null) return;
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摘要：Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set toNULL. Initially, all next
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摘要：//按层从左往右检索 //按层从左往右检索 //按层从左往右检索 //按层从左往右检索 public class Solution { public void connect(TreeLinkNode root) { while (root != null) { TreeLinkNode tempC
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摘要：class Solution { public int sumNumbers(TreeNode root) { return sum(root, 0); } private int sum(TreeNode n, int s) { if (n == null) return 0; if (n.rig
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摘要：给定一个二叉树和所要查找的两个节点，找到两个节点的最近公共父亲节点（LCA）。比如，节点5和1的LCA是3，节点5和4的LCA是5。 class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, Tr
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摘要：public class Solution { public TreeNode insertNode(TreeNode root, TreeNode node) { if(root == null){ return node; } if(root.val > node.val){ //这个树里面没有
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