1151 LCA in a Binary Tree (30 分) The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification:…

LCA in a Binary Tree PAT-1151 本题的困难在于如何在中序遍历和前序遍历已知的情况下找出两个结点的最近公共祖先. 可以利用据中序遍历和前序遍历构建树的思路,判断两个结点在根节点的左右子树,依次递归找到最近祖先 import java.util.HashMap; import java.util.Map; import java.util.Scanner; import java.util.TreeMap; /** * @Author WaleGarrett * @Dat…

题目: Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 说明:1)下面有两种实现:递归(Recursive )与非递归(迭代iterati…

题目描述 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.Given any two nodes in a binary tree, you are supposed to find their LCA. 最小共同祖先(LCA)是一棵树中两个节点U和V最深的那个公共父节点.要求给一棵树,以及两个节点,请你…

https://pintia.cn/problem-sets/994805342720868352/problems/1038430130011897856 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are sup…

题目 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one t…

给定先序中序遍历的序列,可以确定一颗唯一的树 先序遍历第一个遍历到的是根,中序遍历确定左右子树 查结点a和结点b的最近公共祖先,简单lca思路: 1.如果a和b分别在当前根的左右子树,当前的根就是最近祖先 2.如果根等于a或者根等于b了,根就是最近祖先:判断和a等还是和b等就行了 3.如果都在左子树上,递归查左子树就可以了.这里找到左子树的边界和根(通过先序中序序列) 4.如果都在右子树上,递归查右子树. 代码1:不建树的做法,参考柳婼blog~ #include<bits/stdc++.h>…

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one test…

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one test…

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one test…

The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one test…

题目: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 说明: 1)递归和非递归实现,其中非递归有两种方法 2)复杂度,时间O(n),空…

根据一棵树的中序遍历与后序遍历构造二叉树. 注意: 你可以假设树中没有重复的元素. 例如,给出 中序遍历 inorder = [9,3,15,20,7] 后序遍历 postorder = [9,15,7,20,3] 返回如下的二叉树: 3 / \ 9 20 / \ 15 7 class Solution { public: TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { if(p…

翻译 给定一个二叉树.返回其兴许遍历的节点的值. 比如: 给定二叉树为 {1. #, 2, 3} 1 \ 2 / 3 返回 [3, 2, 1] 备注:用递归是微不足道的,你能够用迭代来完毕它吗? 原文 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: R…

Source: PAT A1151 LCA in a Binary Tree (30 分) Description: The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their…

题目 The following is from Max Howell @twitter: Google: 90% of our engineers use the sofware you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck of. Now it's your turn to prove that YOU CAN invert a binary tree! Input Speci…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > re…

Invert a binary tree 翻转一棵二叉树 假设有如下一棵二叉树: 4  / \   2    7  / \   / \ 1  3 6  9翻转后: 4     /    \    7     2   / \    / \  9  6  3  1 这里采用递归的方法来处理.遍历结点,将每个结点的两个子结点交换位置即可. 从左子树开始,层层深入,由底向上处理结点的左右子结点:然后再处理右子树 全部代码如下: public class InvertBinaryTree { public…

这道题是LeetCode里的第102道题. 题目要求: 给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 例如: 给定二叉树: [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回其层次遍历结果: [ [3], [9,20], [15,7] ] 解题代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *…

Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > r…

题目: 给定一个二叉树,返回它的中序 遍历. Given a binary tree, return the inorder traversal of its nodes' values. 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,3,2] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? Follow up: Recursive solution is trivial, could you do it iteratively? 解题思路: 百度百科:二叉树的…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example: Input: [1,2,3,null,5,null,4] Output: [1, 3, 4] Explanation: 1 <--- / \ 2 3 <--- \ \ 5 4 <---  …

94. 二叉树的中序遍历 94. Binary Tree Inorder Traversal 题目描述 给定一个二叉树,返回它的 中序 遍历. LeetCode94. Binary Tree Inorder Traversal 示例: 输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,3,2] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? Java 实现 Iterative Solution import java.util.LinkedList; import java…

Level:   Medium 题目描述: Given a binary tree, return the inorder traversal of its nodes' values. 思路分析:   实现一棵二叉树的中序遍历,我们可以用简单的递归方法去实现,也可以使用栈去实现,使用第二种方式时,我们沿着根节点先遍历左子树的左孩子,将它们依次压入栈,知道左孩子为空,弹出栈顶节点,这时记录栈顶节点的值,如果栈顶节点的右孩子不为空,压入栈,如果为空,则栈顶元素继续弹出,重复上述操作,就能获得中序遍…

236. Lowest Common Ancestor of a Binary Tree /** * 基础版 * 给定p,q都是在树中 * 有两种情况: * 1. p和q分布在LCA的两侧 * 2. p和q在同一侧(p是q的祖先 或者是 q是p的祖先) * 先考虑第一种情况,例如: * 1 * / \ * 2 4 * 2和4的LCA是1,向下递归返回时,2返回2,4返回4,1接受到的left和right的返回值都存在值,那么就返回自己 * 也就是说=> 只要找到目标节点,就往上返回该节点 * -…

本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90577042 1102 Invert a Binary Tree (25 分)   The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary…

The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck off. Now it's your turn to prove that YOU CAN invert a binary tree! Input Specif…

题意:根据前序和中序建立树,寻找两个点的LCA. 我在之前的博客中写了关于LCA的多种求法. https://www.cnblogs.com/yy-1046741080/p/11505547.html.  在建树的过程中,建立深度和parent,来寻找LCA. 该题目的数据有一定的欺诈性,它给你结点数据是1-8,如果没有仔细看清题目,那么很有可能定义一个 node tree[10005]的数组,但是题目并没有说数据范围在1-10000内. 经过测试,如果你将范围定义稍微大一点,还是能全过的: 我…

235. Lowest Common Ancestor of a Binary Search Tree Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between…