4. leetcode (binary tree traversal)

binary tree

Binary tree traversal

A full binary tree has only nodes with a degree of 0 or 2, and nodes with a degree of 0 are on the same layer; the depth of a full binary tree is k, and the number of nodes is 2^k - 1; a complete binary tree is not filled except for the
bottom node , the number of nodes in the remaining layers has reached the maximum value, and the nodes in the bottom layer are all concentrated in the leftmost position of the layer; the
priority queue is actually a heap, and the heap is a complete binary tree, while ensuring the order of the parent and child nodes relation.
Binary search tree
Balanced binary search tree (AVL) : It is an empty tree or the absolute value of the height difference between its left and right subtrees does not exceed 1, and both left and right subtrees are a balanced binary tree.
The underlying implementations of map, set, multimap, and multiset in C++ are all balanced binary trees, so the complexity of the addition and deletion operations of map and set is logn, but the underlying implementation of unordered_map and unordered_set is the storage method of
hash table binary tree , chain storage, order The storage array stores the binary tree, the subscript of the parent node is i, the left child i 2 + 1 the right child i 2 + 2; depth-first traversal (front, middle, post-order traversal) breadth-first traversal (level order traversal)

//前序遍历   递归

#include <iostream>
#include <vector>

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

class Solution {
    
    
public:
    void traversal(TreeNode* cur, std::vector<int>& v) {
    
    
        if (cur == NULL) {
    
    
            return;
        }
        //中序、后序遍历调整位置即可
        v.push_back(cur->val);  // 中
        traversal(cur->left, v); // 左
        traversal(cur->right, v); // 右
    }
    std::vector<int> perorderTravelsal(TreeNode* root) {
    
    
        std::vector<int> res;
        traversal(root, res);
        return res;
    }
};

1. leetcode 144 preorder traversal of binary tree

/**
 * 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) {}
 * };
 */

//递归
class Solution {
    
    
public:
    void traversal(TreeNode* root, vector<int>& res) {
    
    
        if (root == NULL) {
    
    
            return;
        }

        res.push_back(root->val);
        traversal(root->left,  res);
        traversal(root->right, res);
    }

    vector<int> preorderTraversal(TreeNode* root) {
    
    
        vector<int> res;
        traversal(root, res);
        return res;
    }
};

//迭代
class Solution {
    
    
public:
    vector<int> preorderTraversal(TreeNode* root) {
    
    
        vector<int> res;
        stack<TreeNode*> st;
        if (root == nullptr) {
    
    
            return res;
        }

        st.push(root);
        while (!st.empty()) {
    
    
            TreeNode* node = st.top();
            st.pop();
            res.push_back(node->val);

            if (node->right) {
    
    
                st.push(node->right);
            }
            if (node->left) {
    
    
                st.push(node->left);
            }
        }
        return res;
    }
};


//迭代 统一
class Solution {
    
    
public:
    vector<int> preorderTraversal(TreeNode* root) {
    
    
        vector<int> result;
        stack<TreeNode*> st;
        if (root != NULL) st.push(root);
        while (!st.empty()) {
    
    
            TreeNode* node = st.top();
            if (node != NULL) {
    
    
                st.pop();
                if (node->right) st.push(node->right);  // 右
                if (node->left) st.push(node->left);    // 左
                st.push(node);                          // 中
                st.push(NULL);
            } else {
    
    
                st.pop();
                node = st.top();
                st.pop();
                result.push_back(node->val);
            }
        }
        return result;
    }
};

2. Leetcode 145 post-order traversal of binary tree

//递归
class Solution {
    
    
public:
    void traversal(TreeNode* root, vector<int>& res) {
    
    
        if (root == nullptr) {
    
    
            return;
        }
        
        traversal(root->left, res);
        traversal(root->right, res);
        res.push_back(root->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
    
    
        vector<int> res;

        traversal(root, res);
        return res;
    }
};

//迭代

class Solution {
    
    
public:
    vector<int> postorderTraversal(TreeNode* root) {
    
    
        vector<int> res;
        stack<TreeNode*> st;
        if (root == nullptr) {
    
    
            return res;
        }

        st.push(root);
        while (!st.empty()) {
    
    
            TreeNode* node = st.top();
            st.pop();
            res.push_back(node->val);

            if (node->left) {
    
    
                st.push(node->left);
            }
            if (node->right) {
    
    
                st.push(node->right);
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};


//迭代 统一

class Solution {
    
    
public:
    vector<int> postorderTraversal(TreeNode* root) {
    
    
        vector<int> result;
        stack<TreeNode*> st;
        if (root != NULL) st.push(root);
        while (!st.empty()) {
    
    
            TreeNode* node = st.top();
            if (node != NULL) {
    
    
                st.pop();
                st.push(node);                          // 中
                st.push(NULL);

                if (node->right) st.push(node->right);  // 右
                if (node->left) st.push(node->left);    // 左

            } else {
    
    
                st.pop();
                node = st.top();
                st.pop();
                result.push_back(node->val);
            }
        }
        return result;
    }
};

3. Inorder traversal of leetcode 94 binary tree

//递归
/**
 * 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* root, vector<int>& res) {
    
    
        if (root == nullptr) {
    
    
            return;
        }

        traversal(root->left, res);
        res.push_back(root->val);
        traversal(root->right, res);
    }
    vector<int> inorderTraversal(TreeNode* root) {
    
    
        vector<int> res;

        traversal(root, res);
        return res;
    }
};

//迭代

class Solution {
    
    
public:
    vector<int> inorderTraversal(TreeNode* root) {
    
    
        vector<int> res;
        stack<TreeNode*> st;
        TreeNode* cur = root;
        while (cur != nullptr || !st.empty()) {
    
    
            if (cur != nullptr) {
    
      //指针访问节点,访问到最底层
                st.push(cur);   //将访问的节点放入栈
                cur = cur->left;   //左
            } else {
    
    
                cur = st.top(); // 弹出的数据就是要处理的数据
                st.pop();
                res.push_back(cur->val);  //中
                cur = cur->right;   //右
            }
        }
        return res;
    }
};

// 迭代 统一
class Solution {
    
    
public:
    vector<int> inorderTraversal(TreeNode* root) {
    
    
        vector<int> res;
        stack<TreeNode*> st;

        if (root != nullptr) {
    
    
            st.push(root);
        }

        while (!st.empty()) {
    
    
            TreeNode* node = st.top();
            if (node != nullptr) {
    
    
                st.pop();   // 将该节点弹出,避免重复操作,下面再将右中左节点添加到栈中
                if (node->right) {
    
      //添加右节点
                    st.push(node->right);
                }
                st.push(node);      //添加中节点
                st.push(nullptr);   //中节点访问过,但是还没处理,加入空节点作为标记
                if (node->left) {
    
    
                    st.push(node->left);   //添加左节点
                } 
            } else {
    
         //只遇到空节点的时候,才将下一个节点放入res
                st.pop();   //将空节点弹出
                node = st.top();   // 重新取出栈中元素
                st.pop();
                res.push_back(node->val);  //加入res
            }
        }
        return res;
    }
};

4. Level order traversal of leetcode 102 binary tree

/**
 * 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<vector<int>> levelOrder(TreeNode* root) {
    
    
        vector<vector<int>> res;
        queue<TreeNode*> que;
        if (root != nullptr) {
    
    
            que.push(root);
        }

        while (!que.empty()) {
    
    
            int size = que.size();
            vector<int> vec;

            for (int i = 0; i < size; i++) {
    
    
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node->val);

                if (node->left) {
    
    
                    que.push(node->left);
                }
                if (node->right) {
    
    
                    que.push(node->right);
                }
            }
            res.push_back(vec);
        }
        return res;
    }
};

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Origin blog.csdn.net/weixin_44847326/article/details/123534809