LeetCode（剑指 Offer）- 68 - I. 二叉搜索树的最近公共祖先

Posted 2022-05-21 放羊的牧码

题目链接：点击打开链接

题目大意：略

解题思路：略

相关企业

• 字节跳动
• Facebook
• 亚马逊（Amazon）
• 谷歌（Google）
• 微软（Microsoft）
• 腾讯（Tenent）
• 快手
• 小米集团
• 苹果（Apple）
• 领英（LinkedIn）

AC 代码

• Java

/**
 * Definition for a binary tree node.
 * public class TreeNode 
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x)  val = x; 
 * 
 */
 
// 解决方案(1)
class Solution 
 
    Stack<TreeNode> pStack = new Stack<>(), qStack = new Stack<>();
 
    Map<TreeNode, Integer> pMap = new HashMap<>(), qMap = new HashMap<>();
 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) 
        dfs(root, p.val, pStack, pMap, 0);
        dfs(root, q.val, qStack, qMap, 0);
        TreeNode res = common();
        return res;
    
 
    TreeNode common() 
        TreeNode res;
        while (!pStack.isEmpty() && !qStack.isEmpty()) 
            TreeNode pe = pStack.peek();
            TreeNode qe = qStack.peek();
            if (pe.val == qe.val) 
                res = pStack.peek();
                pStack.pop();
                qStack.pop();
                return res;
            
 
            if (pMap.get(pe) > qMap.get(qe)) 
                pStack.pop();
             else if (pMap.get(pe) < qMap.get(qe)) 
                qStack.pop();
             else 
                pStack.pop();
                qStack.pop();
            
        
        return null;
    
 
    void dfs(TreeNode node, int val, Stack<TreeNode> stack, Map<TreeNode, Integer> map, int l) 
        if (node == null) 
            return;
        
        if (node.val == val) 
            stack.push(node);
            map.put(node, l);
         else if (node.val > val) 
            stack.push(node);
            map.put(node, l);
            dfs(node.left, val, stack, map, l + 1);
         else 
            stack.push(node);
            map.put(node, l);
            dfs(node.right, val, stack, map, l + 1);
        
    

 
// 解决方案(2)
class Solution 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) 
        if(p.val > q.val)  // 保证 p.val < q.val
            TreeNode tmp = p;
            p = q;
            q = tmp;
        
        while(root != null) 
            if(root.val < p.val) // p,q 都在 root 的右子树中
                root = root.right; // 遍历至右子节点
            else if(root.val > q.val) // p,q 都在 root 的左子树中
                root = root.left; // 遍历至左子节点
            else break;
        
        return root;
    

 
// 解决方案(3)
class Solution 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) 
        if(root.val < p.val && root.val < q.val)
            return lowestCommonAncestor(root.right, p, q);
        if(root.val > p.val && root.val > q.val)
            return lowestCommonAncestor(root.left, p, q);
        return root;
    

• C++

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) 
 * ;
 */

// 解决方案(1)
class Solution 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
        if(p->val > q->val)
            swap(p, q);
        while(root != nullptr) 
            if(root->val < p->val) // p,q 都在 root 的右子树中
                root = root->right; // 遍历至右子节点
            else if(root->val > q->val) // p,q 都在 root 的左子树中
                root = root->left; // 遍历至左子节点
            else break;
        
        return root;
    
;
 
// 解决方案(2)
class Solution 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
        if(root->val < p->val && root->val < q->val)
            return lowestCommonAncestor(root->right, p, q);
        if(root->val > p->val && root->val > q->val)
            return lowestCommonAncestor(root->left, p, q);
        return root;
    
;