剑指offer--面试题50：树中两个结点的最低公共祖先

来源：互联网发布：网络推广属于什么岗位编辑：程序博客网时间：2024/05/01 08:39

1. 二叉排序树中求解两个结点的最低公共祖先

python实现：

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def lowestCommonAncestor(self, root, p, q):        """        :type root: TreeNode        :type p: TreeNode        :type q: TreeNode        :rtype: TreeNode        分析：二叉排序树，设p,q的最小公共祖先是r，那么应满足p.val<=r.val and q.val >=r.val，即p,q应在r的两侧        所以从根节点开始按照某种遍历搜索，检查当前节点是否满足这种关系，如果满足则返回。如果不满足，例如        p,q都大于r,那么继续搜索r的右子树        """        if root:            if root.val > p.val and root.val > q.val:                return self.lowestCommonAncestor(root.left, p, q)            elif root.val < p.val and root.val < q.val:                return self.lowestCommonAncestor(root.right, p, q)            else:                return root        return None

2.普通二叉树中求解两个结点的最低公共祖先

Python实现：

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def lowestCommonAncestor(self, root, p, q):        """        :type root: TreeNode        :type p: TreeNode        :type q: TreeNode        :rtype: TreeNode        """        if root is None or p is None or q is None:            return None        path1 = []        self.getPath(root, p, path1)        path2 = []        self.getPath(root, q, path2)        #if len(path1) == 0 or len(path2) == 0:        #    return None        result = None        for i in range(min(len(path1), len(path2))):            if path1[i] == path2[i]:                result = path1[i]        return result            def getPath(self, root, targetNode, path):        #当找到targetNode后，targetNode后面的节点就不再append        if len(path)>0 and path[-1] == targetNode:            return        if root:            path.append(root)            self.getPath(root.left, targetNode, path)            self.getPath(root.right, targetNode, path)            #如果还没有找到targetNode，就pop，不用担心路径上的点会pop掉，因为targetNode加入路径后，他路径上的祖先节点才会执行            #下面这个if判断（递归栈的顺序）            if len(path)>0 and path[-1] != targetNode:                path.pop()#######

0 0