LeetCode.235(236) Lowest Common Ancestor of a Binary Search Tree && II

题目235：

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.
分析：/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
//给定二叉查找树，找出两个节点最低公共祖先
//思路：因为符合规则，左子树小于根节点，右子树大于根节点。所以递归判断判断两个节点是否在同一边。
if(root.val>p.val&&root.val>q.val){
//左子树
return lowestCommonAncestor(root.left,p,q);
}else if(root.val<p.val&&root.val<q.val){
return lowestCommonAncestor(root.right,p,q);
}else{
//两边的各有一个
return root;
}

}
}题目236：

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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).” _______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
For example, the lowest common ancestor (LCA) of nodes

5

and

1

is

3

. Another example is LCA of nodes

5

and

4

is

5

, since a node can be a descendant of itself according to the LCA definition.
分析：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
//给定二叉树（不存在规则），找出其最后公共祖先节点
//思路：找出包含两个目标节点路径，最后匹配两条路径,最后一个公共节点就是目标值
if(root==null||root==p||root==q){
//只要匹配其中一个，则为最后的公共节点
return root;
}
//递归查找左边
TreeNode left=lowestCommonAncestor(root.left,p,q);
TreeNode right=lowestCommonAncestor(root.right,p,q);
return left==null?right:right==null?left:root;
}
}