Lowest Common Ancestor of a Binary Tree"/>
Leetcode 236 Lowest Common Ancestor of a Binary Tree
Leetcode 236 Lowest Common Ancestor of a Binary Tree
题目描述
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]
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
Note:
- All of the nodes’ values will be unique.
- p and q are different and both values will exist in the binary tree.
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链接:
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思路解析
-
思路一(递归
root结点有三种可能:
- root is None
- root is p
- root is q
遇到这三种,直接返回root即可;但如果root不是上述三种情况,则有三种情况:
- p、q、LCA均在当前root的左子树上
- p、q、LCA均在当前root的右子树上
- p、q分别在当前root的左右子树上,LCA为当前root
代码如下:
# 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 root is p or root is q:return rootleft = self.lowestCommonAncestor(root.left, p, q)right = self.lowestCommonAncestor(root.right, p, q)if not left:return rightif not right:return leftreturn root
时间复杂度最坏为 O ( N ) O(N) O(N),空间复杂度最坏为 O ( N ) O(N) O(N)
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Leetcode 236 Lowest Common Ancestor of a Binary Tree
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