二叉树的重要遍历问题

PS:本文主要包括二叉树的前序，中序，后序，层序，垂序遍历，使用cpp语言。 

前序遍历

struct TreeNode {
        int val;
        TreeNode* left;
        TreeNode* right;
        TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
        vector<int> preorderTraversal(TreeNode* root) {
               stack<TreeNode*> sta;
               vector<int> res; // 返回的数组
               if (root == nullptr) return res; // 判断根节点,空树，直接返回
               sta.push(root);
               while (!sta.empty()) {
                       TreeNode* tmp = sta.top();
                       sta.pop();
                       if (tmp->right != nullptr) sta.push(tmp->right);
                       if (tmp->left != nullptr) sta.push(tmp->left);
                       res.push_back(tmp->val);
               }
               return res;
        }
};

中序遍历

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack <TreeNode*> sta;
        vector <int> res;
        if (root == nullptr) return res;  // 空树， 直接返回res
        TreeNode* cur = root;
        while (cur != nullptr || !sta.empty) {
            // 先遍历左孩子
            if (cur != nullptr ){
                sta.push(cur);
                cur = cur->left;
            }
            else {
                // 再 是 根结点
                TreeNode* temp = sta.top();
                sta.pop();  // 访问完根节点后就可以出栈，左孩子和根节点都访问完
                res.push_back(temp->val);
                // 最后右孩子
                cur = temp->right;
            }
        }
        return res;
    }
};

后序遍历

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

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

// 第一种写法
class Solution2 {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> sta;
        vector<int> res ;
        if (root == nullptr) return res;
        TreeNode* cur = root;  
        sta.push(cur);
        while(!sta.empty()){
            TreeNode* temp = sta.top();
            res.push_back(temp->val);
            sta.pop();
            if (temp->left) sta.push(temp->left);
            if (temp->right) sta.push(temp->right);


        }
        reverse(res.begin(),res.end());
        return res;   
    }
};


// 第二种方法 (重点理解)
vector<int> PostOrderLoop(TreeNode* root)
{
    stack<TreeNode*> sta;
    vector<int> ans;
    TreeNode* cur = root;
    TreeNode* pre = nullptr;
    while (cur || !sta.empty())
    {
        while (cur)
        {
            sta.push(cur);
            cur = cur->left;
        }
        cur = sta.top();
        //若右节点已经访问过或者没有右节点  则输出该节点值
        if (cur->right == nullptr || pre == cur->right) {
            sta.pop();
            ans.push_back(cur->val);
            pre = cur;
            cur = nullptr;
        }
        else {
            cur = cur->right;
            pre = nullptr;
        }
    }
    return ans;
}

层序遍历

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

//Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

//  BFS
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if (root == nullptr) return res;
        queue<TreeNode*> que;
        que.push(root);
        while (!que.empty()) {
            int size = que.size();
            vector<int> level;
            while (size--) {
                TreeNode* temp = que.front();
                que.pop();
                level.push_back(temp->val);
                if (temp->left != nullptr) que.push(temp->left);
                if (temp->right != nullptr) que.push(temp->right);
            }
            res.push_back(level);
        }
        return res;
    }
};

垂序遍历

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
         map<int, vector<pair<int, int>>> tree;
         void dfs(TreeNode* root, int x, int y) {
                if (root == nullptr)  return;
                tree[x].push_back(make_pair(y, root->val));
                dfs(root->left, x - 1, y + 1);   // 将节点进行给予坐标处理
                dfs(root->right, x + 1, y + 1);
         }
         vector<vector<int>> verticalTraversal(TreeNode* root) {
                vector<vector<int>> res;
                if (root == nullptr) return res;
                dfs(root, 0, 0);
                for (auto& node : tree) {    // 由于要修改node的值，需要加引用
                        sort(node.second.begin(), node.second.end());
                        vector<int> tmp;
                        for (auto t : node.second) {
                               tmp.push_back(t.second);   // 获取val
                        }
                        res.push_back(tmp);
                }
                return res;
         }
};