# 2, Binary tree

## 1. General

• Definition: constraints are imposed on general trees:

• Each node can have at most two subtrees, that is, there are no binary trees   Degree greater than 2   Node of

• Subtree has   Left right order   Division of

• There are 5 forms: • Full binary tree and complete binary tree (delete nodes from right to left at the bottom of full binary tree)

## 2. Important characteristics

• Binary tree, in the second   i   Layers have at most   2i-1   Nodes
• Depth is   k   A binary tree has at most   2k-1   Nodes
• Height (or depth) is   K   A complete binary tree of at least   2k-2   individual   leaf node
• Of nonempty binary tree   Leaf node number   Equal to 2 degrees   The number of nodes plus 1, that is, n0 = n2 + 1

n1 of a complete binary tree can only be 0 or 1

• One degree is   m   If the node with degree 1 is n1, the node with degree 2 is n2,..., and the number of nodes with degree m is nm, then the number of leaf nodes is n0**= 1 + n2****+ 2n3 * * * * +... + (m-1)nm**
• Complete binary tree with n nodes, depth   log2n + 1
• Number property: complete binary tree with n nodes (its depth is   log2n + 1), number each node from top to bottom and from left to right (1~n), then: if i is the number of a node:
• If I is not equal to 1, the parent node number of a is:   ⌊ i/2 ⌋
• If   2I ≤ n, then the left child number of a is   2i； If * * 2I > n * *, then a**   No left child * *;
• If * * 2I + 1 ≤ n * *, the right child number of a is**   2i + 1**； If**   2I + 1 > n * *, then a**   No right child * *;

# 3, Storage structure of binary tree

## 1. Sequential storage

• use   array   To store data elements
• From the storage point of view, this sequential storage structure is only applicable to   Complete binary tree

Because in the worst case, a depth of   k   And only   k   A single tree with two nodes (there is no node with degree 2 in the tree), but it needs a length of   2k-1   A one-dimensional array of.

## 2. Chain storage

• with   Linked list   Store data elements and the relationship between data elements. # 4, Traversal of binary tree

## 1. Determine the binary tree from the traversal sequence

• from   First order and middle order can be determined
• from   Post order and middle order can be determined (but note that the last post order is the root and the next is the right subtree root)
• from   Hierarchy and middle order can be determined

## 2. Estimate the binary tree according to the traversal sequence

• Preamble   Ergodic sequence sum   Post order   Ergodic sequence   identical   Tree: only root node
• Preamble   Ergodic sum   Middle order   ergodic   identical   Binary tree: all nodes have no left subtree (right single branch tree)
• Middle order   Ergodic sum   Post order   ergodic   identical   Binary tree: all nodes have no right subtree (left single branch tree)
• Preorder ergodic sum   Post order   ergodic   contrary   Binary tree: no left subtree or no right subtree (only one leaf node) with a height equal to the number of nodes
• Preamble   Ergodic sum   Middle order   ergodic   contrary   Binary tree: all nodes have no right subtree (left single branch tree)
• Middle order   Ergodic sum   Post order   ergodic   contrary   Binary tree: all nodes have no left subtree (right single branch tree)

## 3. Traversal and tree building code

• Construction of binary tree
• Depth first traversal (first order, middle order and last order)
• Breadth first traversal (first order, second order)
```/* BitTree.java */

package com.java.tree;

import java.util.Queue;

/**
* Created by Jaco.Young.
* 2018-06-13 18:26
*/
public class BitTree {

//Represents the root node of a tree uniquely determined by preorder and inorder
private TreeNode root;

/**
* Methods provided to external calls
* The character array pre represents the preorder traversal sequence, and mid represents the medium order traversal sequence
*/
public void build(char[] pre, char[] mid){
//Assign the root node of the created tree to root
root = buildTree(pre,0, pre.length-1, mid, 0, mid.length-1);
}

/**
* Prerequisite: there are no duplicate elements in the tree
* A method of constructing binary tree from preorder traversal sequence and middle order traversal sequence
* In the process of building a tree, we always divide the sequence into left subtree and right subtree
* lPre,rPre And lMid, rMid, respectively, indicate which part of the preorder and middle order to create a tree
*/
private TreeNode buildTree(char[] pre, int lPre, int rPre, char[] mid, int lMid, int rMid){
//In the preorder traversal sequence, find the root node of the current tree
char root = pre[lPre];

//In the middle order traversal sequence, the position in the middle order is found according to the root node in the first order
int rootIndex = getRootIndex(mid, lMid, rMid, root);

if(rootIndex == -1){
throw new IllegalArgumentException("Illegal Argument!");
}

//Calculate the number of left and right subtrees of the current tree
//Entire middle order sequence: [left subtree (lmid) root (rootindex) right subtree (rMid)]
//Left subtree [lMid,rootIndex-1]
int lNum = rootIndex - lMid; //rootIndex-1 -lMid + 1
//Right subtree [rootIndex+1,rMid]
int rNum = rMid - rootIndex;  //rMid - (rootIndex + 1) + 1

//Start to build the left and right subtrees of the current root node
//First build the left subtree
TreeNode lchild;  //As the root node of the left subtree
//The tree with the current node as the root has no left subtree
if(lNum == 0){
lchild = null;
}else{
//At present, the left subtree of this tree is still a tree. Recursively construct the left subtree of this tree
//Let x be the subscript of the last element of the left subtree in the current tree precedence, then: x - (lpre + 1) = lNum
//Get: x = lPre + lNum
lchild = buildTree(pre, lPre + 1, lPre+lNum, mid, lMid, rootIndex - 1);
}

//Constructing right subtree
TreeNode rchild;
if(rNum == 0){
rchild = null;
}else{
//The right subtree of the current node still contains many nodes, and its right subtree needs to be constructed recursively
rchild = buildTree(pre, lPre + lNum + 1, rPre, mid, rootIndex + 1, rMid);
}

//Construct a complete binary tree
return new TreeNode(root,lchild,rchild);
}

//In the middle order traversal sequence, the position in the middle order is found according to the root node in the first order
private int getRootIndex(char[] mid, int lMid, int rMid, char root) {
for(int i = lMid; i <= rMid; i++){
if(mid[i] == root){
return i;
}
}
return -1;
}

//Structure of each node of binary tree
private class TreeNode{
//Data stored in nodes
char item;
//Point to left child node

> **Java Network disk: pan.baidu.com/s/1MtPP4d9Xy3qb7zrF4N8Qpg
> Extraction code: 2 p8n**

### last

After all, I've been working for so long. Besides tuhu's round, I also interviewed many big factories in July and August. For example, Ali, hungry, meituan and Didi, the interview process will not be written in this article one by one. I'll sort out a detailed interview process and some questions you want to know

### Meituan interview experience
![Meituan interview](https://img-blog.csdnimg.cn/img_convert/120428b3297bdceb312b567878d65d5e.png)
Byte interview experience
![Byte interview](https://img-blog.csdnimg.cn/img_convert/0dc1f994d7c60586500c1b3e58f38e38.png)
Rookie interview experience
![Rookie interview](https://img-blog.csdnimg.cn/img_convert/1c48323a4391f12b5f2c4eb96390de0d.png)
Ant gold interview experience
![Ant gold suit](https://img-blog.csdnimg.cn/img_convert/e91964dd4ec9dc679a77da2d0da2bb0e.png)
Vipshop interview experience
![Vipshop](https://img-blog.csdnimg.cn/img_convert/8f24d4bf51040f428ec2f4763d1754a5.png)

>Due to limited space, pictures and texts cannot be sent out in detail
>
>**[CodeChina Open source project: [first tier big factory] Java Analysis of interview questions+Core summary learning notes+Latest explanation Video]](

)**

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Ant gold interview experience
[External chain picture transfer...(img-mKzl7NVe-1631407720282)]
Vipshop interview experience
[External chain picture transfer...(img-x4YTi4zo-1631407720282)]

>Due to limited space, pictures and texts cannot be sent out in detail
>
>**[CodeChina Open source project: [first tier big factory] Java Analysis of interview questions+Core summary learning notes+Latest explanation Video]](

)**

```

Posted on Fri, 19 Nov 2021 23:52:47 -0500 by popsiclesph