模拟系统栈的二叉树非递归遍历

最新推荐文章于 2022-06-28 16:52:58 发布

原创最新推荐文章于 2022-06-28 16:52:58 发布 · 195 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#二叉树 #算法 #leetcode #c++

数据结构与算法同时被 2 个专栏收录

38 篇文章

订阅专栏

leetcode题解

35 篇文章

订阅专栏

本文详细介绍二叉树的前序、中序和后序遍历算法，采用模拟系统栈的非递归实现方式，帮助读者深入理解遍历过程与递归及栈的关系。

来自leetcode原题

二叉树前序遍历、中序遍历、后续遍历

声明：

有别于教科书的经典非递归实现方式，本文采用模拟系统栈的方式实现非递归，目的是有助于理解递归与栈的紧密关系。并且，三种遍历的代码形式类似，而经典非递归在后序遍历部分较之前两者有很大差异且不易理解。

前序遍历

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

struct Commend{
    string s; //run,print
    TreeNode* node;
    Commend(string s, TreeNode* node): s(s), node(node){}
};

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> record;
        if(!root)
            return record;

        stack<Commend> stack_Tree;
        stack_Tree.push(Commend("run", root));

        while(!stack_Tree.empty()){
            Commend commend = stack_Tree.top();
            stack_Tree.pop();
            if(commend.s == "print"){
                record.push_back(commend.node->val);
            }
            else{
                assert(commend.s == "run");
                if(commend.node->right)
                    stack_Tree.push(Commend("run", commend.node->right));
                if(commend.node->left)
                    stack_Tree.push(Commend("run", commend.node->left));
                stack_Tree.push(Commend("print", commend.node));
            }
        }

        return record;   
    }
};

中序遍历

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

struct Commend{
    string s;
    TreeNode* node;
    Commend(string s, TreeNode* node) : s(s), node(node) {}
};

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> record;
        if(!root){
            return record;
        }

        stack<Commend> stack_Tree;
        stack_Tree.push(Commend("run", root));

        while(!stack_Tree.empty()){
            Commend commend = stack_Tree.top();
            stack_Tree.pop();

            if(commend.s == "print"){
                record.push_back(commend.node->val);
            }
            else{
                assert(commend.s == "run");
                if(commend.node->right)
                    stack_Tree.push(Commend("run", commend.node->right));
                stack_Tree.push(Commend("print", commend.node));
                if(commend.node->left)
                    stack_Tree.push(Commend("run", commend.node->left));
            }
        }
        
        return record;
    }
};

后序遍历


struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

struct Commend{
    string s;
    TreeNode* node;
    Commend(string s, TreeNode* node) : s(s), node(node) {}
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> record;
        if(!root){
            return record;
        }

        stack<Commend> stack_Tree;
        stack_Tree.push(Commend("run", root));

        while(!stack_Tree.empty()){
            Commend commend = stack_Tree.top();
            stack_Tree.pop();

            if(commend.s == "print"){
                record.push_back(commend.node->val);
            }
            else{
                assert(commend.s == "run");
                stack_Tree.push(Commend("print", commend.node));
                if(commend.node->right)
                    stack_Tree.push(Commend("run", commend.node->right));
                if(commend.node->left)
                    stack_Tree.push(Commend("run", commend.node->left));
            }
        }
        
        return record;
    }
};