數據結構--樹（創建、遍歷）

目前正準備2020屆秋招 算法工程師，複習數據結構！

發現樹遍歷能夠達到O(n)時間複雜度，O(1)空間複雜度（Morris遍歷），趕忙學習了一波。並複習了創建樹、遍歷樹的一些基本操做。

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
struct TreeNode{
    int val;
    TreeNode *left, *right;
    TreeNode(int x): val(x), left(NULL), right(NULL){}
}; 

//初始化樹 
TreeNode* init_tree1(TreeNode *root, int val){
    if(root==NULL){
        root = new TreeNode(val);
        return root;
    }
    if(val<root->val){
        root->left = create_tree(root->left, val); 
    }else{
        root->right = create_tree(root->right, val); 
    }
    
    return root;
}

//初始化樹 
void init_tree2(TreeNode *&root){
    string data;
    cin>>data;
    if(data=="#"){
        root=NULL;
    }else{
        root = new TreeNode(stoi(data));
        init_tree(root->left);
        init_tree(root->right);
    }
}
//中序遍歷，時間複雜度O(n)，空間複雜度O(1) 
void MorrisTraversal(TreeNode *root){
    TreeNode *cur = root, *prev = NULL;
    while(cur!=NULL){
        //1. 若是該節點左子樹爲空，則輸出
        if(cur->left==NULL){
            cout<<cur->val<<" ";
            cur = cur->right;   //指向父親節點 
        }else{
            prev = cur->left;
            while(prev->right!=NULL && prev->right!=cur){
                prev = prev->right;              //找當前節點左子樹最右的節點 
            }
            if(prev->right==NULL){
                prev->right = cur;
                cur = cur->left;
            
            }else if(prev->right==cur){
                //說明該節點直接前驅已訪問，則訪問該節點 
                cout<<cur->val<<" ";
                //下一步指向該節點右子樹，繼續進行
                cur = cur->right; 
                prev->right = NULL; //恢復爲原來節點結構 
            }
        } 
        
    }
    cout<<endl;
}

//中序遍歷，時間複雜度O(n)，空間複雜度O(n) 
void inorder(TreeNode *root){
    if(root){
        inorder(root->left);
        cout<<root->val<<" "; 
        inorder(root->right);     
    }
}

//層次遍歷 
void levelOrder(TreeNode* root){
    queue<TreeNode*> q;
    q.push(root);
    while(!q.empty()){
        int size = q.size();
        for(int i=0;i<size;i++){
            TreeNode *temp = q.front();
            q.pop();
            cout<<temp->val<<" ";
            if(temp->left) q.push(temp->left);
            if(temp->right) q.push(temp->right);
        }
        cout<<endl;
    }
}
int main(){
    TreeNode *root=NULL;
    int s[] = {4,2,6,1,3,5};
    for(int i=0;i<6;i++)
        root = init_tree1(root, s[i]);
    //init_tree2(root);
    levelOrder(root);
    
    cout<<"Morris results: "<<endl;
    MorrisTraversal(root);
    
    cout<<"\nRecursive results:"<<endl;
    inorder(root);
    return 0;
}