LeetCode题目练习-二叉树&二叉搜索树

二叉树遍历

typedef struct TreeNode
{
    int data;
    TreeNode * left;
    TreeNode * right;
    TreeNode * parent;
}TreeNode;
 
void pre_order(TreeNode * Node)
{
    if(Node != NULL)
    {
        printf("%d ", Node->data);
        pre_order(Node->left);
        pre_order(Node->right);
    }
}
void middle_order(TreeNode *Node) {
    if(Node != NULL) {
        middle_order(Node->left);
        printf("%d ", Node->data);
        middle_order(Node->right);
    }
}
void after_order(TreeNode *Node) {
    if(Node != NULL) {
        after_order(Node->left);
        after_order(Node->right);
        printf("%d ", Node->data);
    }
}

验证二叉搜索树

Given a binary tree, determine if it is a valid binary search tree
(BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the
node’s key. The right subtree of a node contains only nodes with keys
greater than the node’s key. Both the left and right subtrees must
also be binary search trees.

Example 1:
2
/ \
1 3
Input: [2,1,3] Output: true

• 算法一：中序遍历这棵树，如果是二叉搜索树，结果应该为升序。O（n）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<int> vec;
    int flag = 1;
public:
    bool isValidBST(TreeNode* root) {
        middle_order(root);
        return flag;
    }
    void middle_order(TreeNode *Node) {
        if(Node != NULL) {
            middle_order(Node->left);
            if(vec.size() == 0||Node->val > vec.back()){
                vec.push_back(Node->val);
            }else {flag = 0;}
            middle_order(Node->right);
    }
}
};

我这个做法里面有flag，不知道好不好，执行用时 :16 ms, 在所有 C++ 提交中击败了77.06%的用户内存消耗 :21.5 MB, 在所有 C++ 提交中击败了5.16%的用户。内存消耗的有点多。应该有改进的地方。

copy一个更好的写法，24 ms， 20.6 MB ，代码简洁一点。

class Solution {
public:
    long x = LONG_MIN;
    bool isValidBST(TreeNode* root) {
        if(root == NULL){
            return true;
        }
        else{
            if(isValidBST(root->left)){
                if(x < root->val){
                    x = root->val;
                    return isValidBST(root->right);
                }
            }
        }
        return false;
    }
};

• 算法二：递归(copy的代码)

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return helper(root,LONG_MIN,LONG_MAX);
    }
    bool helper(TreeNode* cur,long lt,long rt){
        if(cur==NULL) return true;
        if(cur->val<=lt||cur->val>=rt) return false;
        if(helper(cur->left,lt,cur->val)&&helper(cur->right,cur->val,rt)) return true;
        return false;
    }
};

作者：24shi-01fen-_00_01
链接：https://leetcode-cn.com/problems/validate-binary-search-tree/solution/liang-chong-fang-fa-di-gui-zhong-xu-bian-li-by-24s/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

最近公共祖先

Given a binary search tree (BST), find the lowest common ancestor
(LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common
ancestor is defined between two nodes p and q as the lowest node in T
that has both p and q as descendants (where we allow a node to be a
descendant of itself).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

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以节点3和0为例

• 算法一：如果有parent指针，3和0第一个相同的parent节点即为所求
• 算法二：从根节点遍历到0和3，记录他们的路径。如620、6243，第一个分叉的节点2即为所求。O(N)
• 算法三：递归O(N)：注意这个是二叉搜索树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    if(root==NULL) return NULL;
    if(root==p||root==q) return root;//找到了

    if(root->val>p->val&&root->val>q->val) return lowestCommonAncestor(root->left,p,q);
    if(root->val<p->val&&root->val<q->val) return lowestCommonAncestor(root->right,p,q);
    
    return root;
}