題目
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
當其中一個數(shù)是另一個的孩子時孤澎，那個父親節(jié)點就是所求鳖宾。否則，它們分別在最低共同祖先節(jié)點兩側(cè)木张。
代碼
# 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]
