2021.2.1 刷题(二叉树DFS遍历)

1.中序遍历(左-中-右)
题目链接：https://leetcode-cn.com/problems/binary-tree-inorder-traversal/
方法一：递归法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void traversal(TreeNode *cur, vector<int> &result)
    {
        if(cur == nullptr )
            return;
        traversal(cur->left, result);
        result.push_back(cur->val);
        traversal(cur->right, result);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};

方法二:栈-迭代法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode * > st;
        vector<int> result;
        TreeNode * cur = root;  //定义一个指针访问
        while(cur != nullptr || !st.empty())
        {
            if(cur != nullptr)
            {
                st.push(cur);
                cur = cur->left;
            }else{
                cur = st.top();
                result.push_back(cur->val);
                st.pop();
                cur = cur->right;
            }
        }
        return result;
    }
};

2.前序遍历(中-左-右)
题目链接：https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
迭代法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> st;
        vector<int> result;
        st.push(root);
        while(!st.empty())
        {
            TreeNode* p = st.top();
            st.pop();
            if(p != nullptr) result.push_back(p->val);
            else continue;
            st.push(p->right); //栈先进后出
            st.push(p->left);
        }
        return result;
    }
};

3.后序遍历（左-右-中）
题目链接：https://leetcode-cn.com/problems/binary-tree-postorder-traversal/
迭代法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
 
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> st;
        vector<int> result;
        TreeNode *p;
        //遍历顺序  中-右-左
        st.push(root);
        while(!st.empty())
        {
            p = st.top();
            st.pop();
            if(p != nullptr) result.push_back(p->val);
            else continue;
            st.push(p->left); //栈先进后出
            st.push(p->right);
        }
        //翻转成左-右-中
        reverse(result.begin(), result.end());
        return result;
    }
};