难度: 简单(Easy)

题目

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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).”

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

思路

当其中一个数是另一个的孩子时，那个父亲节点就是所求。否则，它们分别在最低共同祖先节点两侧。

代码

# 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 visit(self, root, cur_path, pathes, p, q):
        if root == None:
            return
        cur_path.append(root)
        if root==p:
            pathes.append(cur_path[:])
        if root==q:
            pathes.append(cur_path[:])
        self.visit(root.left, cur_path, pathes, p, q)
        self.visit(root.right, cur_path, pathes, p, q)
        cur_path.pop()

    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if root == None:
            return None
        pathes = []
        self.visit(root, [], pathes, p, q)
        i = 0
        while i < min(len(pathes[0]),len(pathes[1])):
            if pathes[0][i] != pathes[1][i]:
                break
            i += 1
        return pathes[0][i-1]