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 thedefinition 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 allowa node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes2and8is6. Another example is LCA of nodes2and4is2, since a node can be a descendant of itself according to the LCA definition.

二叉搜索树的特点是左<根<右，所以根节点的值一直都是中间值，大于左子树的所有节点值，小于右子树的所有节点值，那么我们可以做如下的判断，如果根节点的值大于p和q之间的较大值，说明p和q都在左子树中，那么此时我们就进入根节点的左子节点继续递归，如果根节点小于p和q之间的较小值，说明p和q都在右子树中，那么此时我们就进入根节点的右子节点继续递归，如果都不是，则说明当前根节点就是最小共同父节点，直接返回即可

recursion

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return None
        # 在左子树
        if root.val > max(p.val,q.val):
            return self.lowestCommonAncestor(root.left, p, q)
        elif root.val < min(p.val, q.val):
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root

递代

用个while循环来代替递归调用即可，然后不停的更新当前的根节点，也能实现同样的效果：

while的终止条件就是：root大于p，q的值或者都小于p，q的值，也就是他们的差都大于0或者都小于0（不等于0，如果等于0还是返回root）。如果root的值介于p，q之间或者等于其中一个值则停止循环。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """

        while (root.val - p.val)*(root.val - q.val) > 0:
            if root.val > max(p.val,q.val):
                root = root.left
            elif root.val < min(p.val,q.val):
                root = root.right
        return root