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leetcode-235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先_lowest common ancestor of a binary search tree giv

lowest common ancestor of a binary search tree given a binary search tree (b

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.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

 

Note:

  • All of the nodes' values will be unique.
  • p and q are different and both values will exist in the BST.

给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。

百度百科最近公共祖先的定义为:“对于有根树 T 的两个结点 p、q,最近公共祖先表示为一个结点 x,满足 x 是 p、q 的祖先且 x 的深度尽可能大(一个节点也可以是它自己的祖先)。”

例如,给定如下二叉搜索树:  root = [6,2,8,0,4,7,9,null,null,3,5]

 

示例 1:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
输出: 6 
解释: 节点 2 和节点 8 的最近公共祖先是 6

示例 2:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。

 

说明:

  • 所有节点的值都是唯一的。
  • p、q 为不同节点且均存在于给定的二叉搜索树中。

思路:二叉搜索树有个性质 就是 左 子节点值<根节点值<右子节点值。那只需要判断 根节点和两个子节点个关系即可。若根节点值大于两个子节点值,那就说明两个节点是 根节点的左子节点,那就让根节点等于它的左子节点。若根节点小于两个子节点值,就说明两个子节点是根节点的右子节点,就让根节点指向 根节点右子节点。否则就说明大于 左子节点小于 右子节点。

递归方法:

  1. /**
  2. * Definition for a binary tree node.
  3. * struct TreeNode {
  4. * int val;
  5. * TreeNode *left;
  6. * TreeNode *right;
  7. * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  8. * };
  9. */
  10. class Solution {
  11. public:
  12. TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
  13. if(!root) return NULL;
  14. if(root->val > max(p->val,q->val))
  15. return lowestCommonAncestor(root->left,p,q);
  16. else if(root->val < min(p->val,q->val))
  17. return lowestCommonAncestor(root->right,p,q);
  18. else return root;
  19. }
  20. };

 循环:

  1. /**
  2. * Definition for a binary tree node.
  3. * struct TreeNode {
  4. * int val;
  5. * TreeNode *left;
  6. * TreeNode *right;
  7. * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  8. * };
  9. */
  10. class Solution {
  11. public:
  12. TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
  13. if(!root) return NULL;
  14. while(true)
  15. {
  16. if(root->val > max(p->val,q->val))
  17. root = root->left;
  18. else if(root->val < min(p->val,q->val))
  19. root = root->right;
  20. else break;
  21. }
  22. return root;
  23. }
  24. };

 

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