235. Lowest Common Ancestor of a Binary Search Tree

题目：

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.

链接：　　http://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

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