【二叉树】刷题 / 复习 记录

二叉树的前/中/后序遍历

https://leetcode.cn/problems/binary-tree-preorder-traversal/

前序遍历：根 -> 左 -> 右

中序遍历：左 -> 根 -> 右

后序遍历：左 -> 右 -> 根

前：

/**
 * 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 dfs(TreeNode *root, vector<int> &ans) {
        if(root == NULL) return;
        ans.push_back(root->val);
        dfs(root->left, ans);
        dfs(root->right, ans);
    }

    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        dfs(root, ans);
        return ans;
    }
};

中：

    void dfs(TreeNode *root, vector<int> &ans) {
        if(root == NULL) return;
        dfs(root->left, ans);
        ans.push_back(root->val);
        dfs(root->right, ans);
    }

后：

    void dfs(TreeNode *root, vector<int> &ans) {
        if(root == NULL) return;
        dfs(root->left, ans);
        dfs(root->right, ans);
        ans.push_back(root->val);
    }

653. Two Sum IV - Input is a BST

/**
 * 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:
    map<int, int> mp;

    bool findTarget(TreeNode* root, int k) {
        if(root == NULL) return 0;
        mp[root->val]++;
        if(mp[k - root->val]) {
            if(k == 2 * root->val && mp[root->val] >= 2 || k != 2 * root->val) return 1;
        }
        bool f1 = 0, f2 = 0;
        if(root->left) {
            f1 = findTarget(root->left, k);
        }   
        if(root->right) {
            f2 = findTarget(root->right, k);
        }
        return f1 || f2;
    }
};