LeetCode: Lowest Common Ancestor of a Binary Search Tree 解题报告

https://leetcode.com/submissions/detail/32662938/

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Leetcode又出新题啦

SOLUTION 1：

使用递归能够轻松解决此问题。对于此题咱们能够分为三种状况讨论：

1. P, Q都比root小，则LCA在左树，咱们继续在左树中寻找LCA

2. P, Q都比root大，则LCA在右树，咱们继续在右树中寻找LCA

3. 其它状况，表示P,Q在root两边，或者两者其一是root，或者都是root，这些状况表示root就是LCA，直接返回root便可。

代码以下：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
12         // 1439.
13         // 1. Two nodes are on left, go left.
14         // 2. Two nodes are on right, go right.
15         // 3. They are in both sides, return root.
16         // if root == null, return null.
17         if (root == null) {
18             return null;
19         }
20         
21         if (p.val < root.val && q.val < root.val) {
22             return lowestCommonAncestor(root.left, p, q);
23         } else if (p.val > root.val && q.val > root.val) {
24             return lowestCommonAncestor(root.right, p, q);
25         } else {
26             return root;
27         }
28     }
29 }

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