LeetCode 235. Lowest Common Ancestor of a Binary Search Tree（二叉搜索树的最低公共祖先）

原题网址：https://leetcode.com/problems/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 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.
方法一：通过递归进行遍历，时间复杂度O(n)，n为节点数。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode lca;
private int find(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || lca != null) return 0;
int found = 0;
if (root.left != null && lca == null) found += find(root.left, p, q);
if (root.right != null && lca == null) found += find(root.right, p, q);
if (root == p) found ++;
if (root == q) found ++;
if (found == 2 && lca == null) lca = root;
return found;
}
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
find(root, p, q);
return lca;
}
}

方法二：利用BST特性，根据当前节点值大小，判断LCA是当前节点，或者在左子树、右子树中。时间复杂度O(log(n))

/**
* 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 (p.val==q.val) return p;
if (p.val > q.val) {
TreeNode t = p;
p=q;
q=t;
}
if (root.val < p.val) return lowestCommonAncestor(root.right, p, q);
if (root.val == p.val) return p;
if (root.val < q.val) return root;
if (root.val == q.val) return q;
return lowestCommonAncestor(root.left, p,q);
}
}