235. Lowest Common Ancestor of a Binary Search Tree

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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.

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  public class Solution {
      public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
          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;
      }
  }

  这个解法简直钛蚌

• 先看recursive。从根节点出发，分别去搜索两个节点，但这里搜索不用瞎搜索，需要利用BST的性质，比较p->val，q->val和root->val的关系，决定往左边搜还是右边搜。如果root->val>p->val同时root->val>q->val,说明p,q都在root的左侧，因此把root=root->left;如果root->val< p->val同时root->val < q->val,说明p,q都在root的右侧，因此把root=root->right;如果root->val>p->val同时root->val< q->val,或root->val< p->val同时root->val> q->val说明p,q在root的两侧，说明root就是LCA.
  ————————————————
  版权声明：本文为CSDN博主「xinqrs01」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
  原文链接：https://blog.csdn.net/xinqrs01/article/details/54999271

235. Lowest Common Ancestor of a Binary Search Tree

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原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11444491.html