题目：
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the : “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.
链接：
题解：
BST求公共祖先。我们把两个节点的值和root进行对比，然后用二分查找的思想进行递归。
Time Complexity - O(logn)， Space Complexity - O(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(root == null || p == null || q == null) return root; if(p.val > root.val && q.val > root.val) return lowestCommonAncestor(root.right, p, q); else if(p.val < root.val && q.val < root.val) return lowestCommonAncestor(root.left, p, q); else return root; }}
二刷：
因为题目给定允许一个节点作为自己的descendant，所以我们只要在root代表的树里同时对p和q进行递归的binary search就可以了。也可以使用迭代。
Java:
Time Complexity - O(logn)， Space Complexity - O(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 (root == null || p == null || q == null) { return root; } if (p.val < root.val && q.val < root.val) { return lowestCommonAncestor(root.left, p, q); } else if (p.val > root.val && q.val > root.val) { return lowestCommonAncestor(root.right, p, q); } else { return root; } }}
使用迭代
Time Complexity - O(logn)， Space Complexity - O(1)
/** * 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 || p == null || q == null) { return root; } TreeNode node = root; while (node != null) { if (p.val < node.val && q.val < node.val) { node = node.left; } else if (p.val > node.val && q.val > node.val) { node = node.right; } else { break; } } return node; }}
三刷:
这里假定 a node can be ancestor of itself，一个节点可以作为它自己的ancestor。我们首先判断边界条件，p = null和q = null时的情况。接下来建立一个node节点 = root。在node 非空的情况下，同时比较node.val和p.val, q.val来决定继续向左子树搜索还是向右子树搜索。假如两个节点分散在node的两边，或者有一个的值等于node，这时node就是lowest common ancestor。
Java:
Time Complexity - O(logn)， Space Complexity - O(1)
/** * 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 == null) { return q; } if (q == null) { return q; } TreeNode node = root; while (node != null) { if (node.val > p.val && node.val > q.val) { node = node.left; } else if (node.val < p.val && node.val < q.val) { node = node.right; } else { return node; } } return root; }}
Update:
/** * 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; TreeNode node = root; while (node != null) { if (node.val < p.val && node.val < q.val) node = node.right; else if (node.val > p.val && node.val > q.val) node = node.left; else return node; } return node; }}
Reference:
http://blog.csdn.net/luckyxiaoqiang/article/details/7518888
http://zhedahht.blog.163.com/blog/static/25411174201081263815813/
https://github.com/julycoding/The-Art-Of-Programming-By-July/blob/master/ebook/zh/03.03.md
http://baozitraining.org/blog/binary-tree-common-anncestor/
http://www.cnblogs.com/felixfang/p/3828915.html
http://www.gocalf.com/blog/least-common-ancestor.html
http://baozitraining.org/blog/ten-rules-for-oop-design/
https://developers.google.com/youtube/v3/code_samples/java
http://www.jiuzhang.com/solutions/lowest-common-ancestor/
https://leetcode.com/discuss/44959/3-lines-with-o-1-space-1-liners-alternatives
