随笔分类 - Tree
easy
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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