LeetCode-235.Lowest Common Ancestor of a Binary Search Tree

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.

根据二叉搜索树的特性，判断结点值的大小，时间复杂度为O(n) ，空间复杂度为O(n)

 1 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {//树 递归 my
 2         if (null==root){
 3             return root;
 4         }
 5         if(p.val<root.val&&q.val<root.val){
 6             return lowestCommonAncestor(root.left,p,q);
 7         }
 8         else if(p.val>root.val&&q.val>root.val){
 9             return lowestCommonAncestor(root.right,p,q);
10         }
11         else{
12             return root;
13         }
14     }

非递归解法，时间复杂度O(n) ，空间复杂度为O(1)

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {//树 mytip
        TreeNode re = root;
        while(null!= re){
            if(p.val<re.val&&q.val<re.val){
                re=re.left;
            }
            else if(p.val>re.val&&q.val>re.val){
                re=re.right;
            }
            else{
                break;
            }
        }
        return  re;
    }

进阶题

LeetCode236 二叉树的最近公共祖先 https://www.cnblogs.com/zhacai/p/10592779.html