二叉树为BST

LCR 193. 二叉搜索树的最近公共祖先

1.1 递归

• 利用BST的性质

1. p == root 或者 q == root ,显然根为公共祖先
2. p < root < q 或者 p > root > q,显然p，q分别位于root的一颗子树上，故根为公共祖先
3. max{p,q} < root ,显然 p 和q 均在root的左子树
4. min{p,q} > root ,显然 p 和q 均在root的右子树

/*** 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 {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(!root)return NULL;if(root->val == p->val || root->val == q->val)return root;if(root->val > q->val && root->val < p->val)return root;if(root->val < q->val && root->val > p->val)return root;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;}
};

1.2 迭代

• 利用后序遍历递归的特点
• 当访问结点p时，此时栈中存储结点 时 自顶向下的祖先

1. 利用后序遍历，获取p和q的祖先序列
2. 因最近公共祖先 所在层数 一定 ≤ min{ stack_p.size() , stack_q.size() }
3. 故先通过出栈让p和q中祖先数量相等，即二者祖先 处于同一层上
4. 当 出栈 到 二者祖先相同时，便为最近公共祖先

/*** 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 {
public:stack<TreeNode*> searchAncestor(TreeNode* root,TreeNode* p){stack<TreeNode*> s;TreeNode* pre;while(root || s.size()){if(root){s.push(root);root = root->left;}else{TreeNode* node = s.top();if(node->val == p->val) break;if(node->right && pre != node->right){root = node->right;}else{s.pop();pre = node;cout << node->val << " ";root = NULL;}}}return s;}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(!root)return NULL;stack<TreeNode*> stack_p = searchAncestor(root,p);stack<TreeNode*> stack_q = searchAncestor(root,q);while(stack_p.size() > stack_q.size()) stack_p.pop();while(stack_q.size() > stack_p.size()) stack_q.pop();while(stack_p.size() && stack_q.size()){if(stack_p.top() == stack_q.top())return stack_p.top();else{stack_p.pop();stack_q.pop();}}return root;}
};

任意二叉树

LCR 194. 二叉树的最近公共祖先

/*** 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 {
public:stack<TreeNode*> searchAncestor(TreeNode* root,TreeNode* p){stack<TreeNode*> s;TreeNode* pre;while(root || s.size()){if(root){s.push(root);root = root->left;}else{TreeNode* node = s.top();if(node->val == p->val) break;if(node->right && pre != node->right){root = node->right;}else{s.pop();pre = node;cout << node->val << " ";root = NULL;}}}return s;}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(!root)return NULL;stack<TreeNode*> stack_p = searchAncestor(root,p);stack<TreeNode*> stack_q = searchAncestor(root,q);while(stack_p.size() > stack_q.size()) stack_p.pop();while(stack_q.size() > stack_p.size()) stack_q.pop();while(stack_p.size() && stack_q.size()){if(stack_p.top() == stack_q.top())return stack_p.top();else{stack_p.pop();stack_q.pop();}}return root;}
};