0.原题:
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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
翻译:
给定一棵二叉树,在树中找到两个给定节点的最近邻的共同祖先(LCA)。
根据Wikipedia上LCA的定义:“最近邻的共同祖先,是在两个节点p和q之间的最近邻的节点,即p和q都是后代(我们允许一个节点成为其自身的后代)。
1.代码:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
if root == None or root == p or root == q:
return root
left = None
right = None
left = self.lowestCommonAncestor(root.left,p,q)
right = self.lowestCommonAncestor(root.right,p,q)
if (left != None) and (right != None):
return root
else:
return (left if(left != None) else(right))
2.思路:
整体思路:采用递归,逐层查找。
step1:设置递归结束条件
如果,查找的子树为空时,即root == 0,需要结束递归,并返回None;
如果,p为当前树的子节点,即root == p,那么p、q的最近邻共同祖先就是p本身,返回p;
如果,q为当前树的子节点,即root == q,那么p、q的最近邻共同祖先就是q本身,返回q;
综上所述:递归结束条件为:root == None or root == p or root == q,返回值都等于 root
step2:查找子树
如果递归没有结束,就说明p和q都在子树之中,需要对子树进行查找。
设置left、right用于存放返回值。
left存放左子树的查找结果,right存放右子树查找结果。
如果,如果在左右子树中,分别查找到了p或者q,那么这时候的root节点就是p、q的最近邻共同祖先。