Lowest Common Ancestor of 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.

思路：找A与B的LCA，有两种可能  1. A或B是另一个点得祖先。比如3，2的LCA就是2。

　　　　　　　　　　　　　　　　2. 一个TreeNode，它的左右子树一个包含A，一个包含B。

　　　这种思路是对于binary tree与binary search tree都适用的，还有一种方法，利用了BST得特点，通过对数值的运算确定了，这个LCA是在root的左子树还是右子树，还是root本身。在leetcode的disscusion中有人提到了。两种的代码如下：

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q）{
        if (root == null || root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        
        if (left != null && right != null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        if (right != null) {
            return right;
        }
        return null;
}

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        while((root.val - p.val) * (root.val - q.val) > 0) {
            root = (root.val - p.val) > 0 ? root.left : root.right;
        }
        return root;
}