[数据结构与算法]二叉树核心算法题（c++）

一、二叉树的遍历

1.前序遍历

144题

/**
 * 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 {
private:
    void recursion(TreeNode *root, vector<int> &ans) {
        if (!root) return;
        ans.push_back(root->val);
        recursion(root->left, ans);
        recursion(root->right, ans);
    }
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        recursion(root, ans);
        return ans;
    }
};

2.中序遍历

/**
 * 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 {
private:
    void recursion(TreeNode* root, vector<int> &ans) {
        if (!root) return;
        recursion(root->left, ans);
        ans.push_back(root->val);
        recursion(root->right, ans);
    }
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ans;
        recursion(root, ans);
        return ans;
    }
};

3.后序遍历

145题

/**
 * 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 {
private:
    void recursion(TreeNode* root, vector<int> &ans) {
        if (!root) return;
        recursion(root->left, ans);
        recursion(root->right, ans);
        ans.push_back(root->val);
    }
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        recursion(root, ans);
        return ans;
    }
};

以上时间复杂度O(n)，平均空间复杂度O(logn)，最坏空间复杂度为O(n)

4. 层序遍历

102题

思路：使用队列，每次将首元素的非空左右子树入队，然后首元素出队，这样对每一层结点进行一次循环，最终就可以得到层次遍历的结果?

/**
 * 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) {
        // 1.准备工作
        queue<TreeNode*> q;
        q.push(root);
        vector<vector<int>> ans;
        vector<int> cur_layer;

        // 2.正式处理
        while (!q.empty()) {
            int sz = q.size();
            cur_layer.clear();
            // 出队并将首元素的非空左右子树压入队列中
            for (int i=0; i<sz; ++i) {
                TreeNode *head = q.front();
                if (head) {
                    cur_layer.push_back(head->val);
                    q.push(head->left);
                    q.push(head->right);
                }
                q.pop();
            }
            if (cur_layer.size() > 0) {
                ans.push_back(cur_layer);
            }      
        }
        return ans;
    }
};

每个结点入队出队一次，时间复杂度O(n)，空间复杂度最坏O(n)

二、二叉查找树/搜索树相关

1. 二叉树的深度

leetcode 104题

/**
 * 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:
    int maxDepth(TreeNode* root) {
        if (!root) return 0;
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
    }
};

时间复杂度O(n)，空间复杂度平均O(height)

2. 验证二叉搜索树

leetcode 98