代码随想录Day18


LeetCode530. 二叉搜索树的最小绝对差

class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        vector<int> array;
        TreeNode* cur = root;
        stack<TreeNode*> s;
        int result = INT_MAX;
        while(cur || !s.empty()){
            while(cur){
                s.push(cur);
                cur = cur->left;
            }
            cur = s.top();
            s.pop();
            array.push_back(cur->val);
            cur = cur->right;
        }
        for(int i = 0; i < array.size() - 1; i++){
            result = min(array[i + 1] - array[i], result);
        }
        return result;
    }
};

LeetCode501.二叉搜索树的众数

class Solution {
public:
    vector<int> findMode(TreeNode* root) {
        TreeNode* cur = root;
        stack<TreeNode*> s;
        vector<int> array, result;
        int numTimes = 0, maxTimes = INT_MIN, nowNum;
        while(cur || !s.empty()){
            while(cur){
                s.push(cur);
                cur = cur->left;
            }
            cur = s.top();
            s.pop();
            array.push_back(cur->val);
            cur = cur->right;
        }
        nowNum = array[0];
        for(int i = 0; i < array.size(); i++){
            if(array[i] == nowNum){
                numTimes++;
            }
            if(array[i] != nowNum){
                if(numTimes == maxTimes){
                    result.push_back(nowNum);
                }else if(numTimes > maxTimes){
                    result.clear();
                    result.push_back(nowNum);
                    maxTimes = numTimes;
                }
                numTimes = 1;
                nowNum = array[i];
            }
            if(i == array.size() - 1){
                if(numTimes == maxTimes){
                    result.push_back(nowNum);
                }else if(numTimes > maxTimes){
                    result.clear();
                    result.push_back(nowNum);
                    maxTimes = numTimes;
                }
            }
        }
        return result;
    }
};

LeetCode236.二叉树最近公共祖先

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == p || root == q || root == NULL)return root;

        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);

        if(left != NULL && right == NULL)return left;
        if(left == NULL && right != NULL)return right;
        if(left != NULL && right != NULL)return root;
        return NULL;
    }
};
### 关于代码随想录 Day04 的学习资料与解析 #### 一、Day04 主要内容概述 代码随想录 Day04 的主要内容围绕 **二叉树的遍历** 展开,包括前序、中序和后序三种遍历方式。这些遍历可以通过递归实现,也可以通过栈的方式进行迭代实现[^1]。 #### 二、二叉树的遍历方法详解 ##### 1. 前序遍历(Pre-order Traversal) 前序遍历遵循访问顺序:根节点 -> 左子树 -> 右子树。以下是基于递归的实现: ```python def preorderTraversal(root): result = [] def traversal(node): if not node: return result.append(node.val) # 访问根节点 traversal(node.left) # 遍历左子树 traversal(node.right) # 遍历右子树 traversal(root) return result ``` 对于迭代版本,则可以利用显式的栈来模拟递归过程: ```python def preorderTraversal_iterative(root): stack, result = [], [] current = root while stack or current: while current: result.append(current.val) # 访问当前节点 stack.append(current) # 将当前节点压入栈 current = current.left # 转向左子树 current = stack.pop() # 弹出栈顶元素 current = current.right # 转向右子树 return result ``` ##### 2. 中序遍历(In-order Traversal) 中序遍历遵循访问顺序:左子树 -> 根节点 -> 右子树。递归实现如下: ```python def inorderTraversal(root): result = [] def traversal(node): if not node: return traversal(node.left) # 遍历左子树 result.append(node.val) # 访问根节点 traversal(node.right) # 遍历右子树 traversal(root) return result ``` 迭代版本同样依赖栈结构: ```python def inorderTraversal_iterative(root): stack, result = [], [] current = root while stack or current: while current: stack.append(current) # 当前节点压入栈 current = current.left # 转向左子树 current = stack.pop() # 弹出栈顶元素 result.append(current.val) # 访问当前节点 current = current.right # 转向右子树 return result ``` ##### 3. 后序遍历(Post-order Traversal) 后序遍历遵循访问顺序:左子树 -> 右子树 -> 根节点。递归实现较为直观: ```python def postorderTraversal(root): result = [] def traversal(node): if not node: return traversal(node.left) # 遍历左子树 traversal(node.right) # 遍历右子树 result.append(node.val) # 访问根节点 traversal(root) return result ``` 而迭代版本则稍复杂一些,通常采用双栈法或标记法完成: ```python def postorderTraversal_iterative(root): if not root: return [] stack, result = [root], [] while stack: current = stack.pop() result.insert(0, current.val) # 插入到结果列表头部 if current.left: stack.append(current.left) # 先压左子树 if current.right: stack.append(current.right) # 再压右子树 return result ``` #### 三、补充知识点 除了上述基本的二叉树遍历外,Day04 还可能涉及其他相关内容,例如卡特兰数的应用场景以及组合问题的基础模板[^2][^4]。如果遇到具体题目,可以根据实际需求调用相应算法工具。 --- ####
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