leetcode--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.

题意：对于一棵BST树，给出树上的两个节点，找出它们的最接近的共同祖先节点。

分类：二叉树

解法1：因为题目给出的是BST树。依据BST树的性质。对于某个节点，其左子树上的全部节点的值都比它小。右子树上的全部节点的值都比它大。

对于根节点root而言。假设p,q两个相比root,一大一小。说明root就是它们的LCA

假设都在比root小。说明它们的LCA在root的左子树上。假设都比root大，说明它们的LCA在root的右子树上。

假设它们有当中一个跟root相等。说明这个就是LCA

依据这个思路，我们非常快写出代码：

/**
* 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==null) return null;
if(p.val<root.val&&root.val<q.val){//假设一左一右
return root;
}else if(p.val>root.val&&root.val>q.val){//假设一左一右
return root;
}else if(p.val<root.val&&root.val>q.val){//假设都在左边，递归查找左子树
return lowestCommonAncestor(root.left,p,q);
}else if(p.val>root.val&&q.val>root.val){//假设都在右边，递归查找右子树
return lowestCommonAncestor(root.right,p,q);
}else if(p.val==root.val){//假设和root相等
return p;
}else if(q.val==root.val){//假设和root相等
return q;
}
return null;
}
}

解法2：解法2跟解法1的思路一样，可是精简了一下代码，除了在左右子树的情况。其余情况我们都能够返回root

/**
* 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==null) return null;
if(p.val<root.val&&root.val>q.val){//假设都在左边。递归查找左子树
return lowestCommonAncestor(root.left,p,q);
}else if(p.val>root.val&&q.val>root.val){//假设都在右边。递归查找右子树
return lowestCommonAncestor(root.right,p,q);
}else{//其余情况
return root;
}
}
}