236 Lowest Common Ancestor of a Binary Tree 二叉树的最近公共祖先
Description:
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:
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.
题目描述:
给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。
百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”
例如，给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]
示例 :
示例 1:
输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
输出: 3
解释: 节点 5 和节点 1 的最近公共祖先是节点 3。
示例 2:
输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
输出: 5
解释: 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。
说明：
所有节点的值都是唯一的。
p、q 为不同节点且均存在于给定的二叉树中。
思路:
递归
- 若 root为空, 或者 root == p or root == q, 直接返回 root, 这时, 要么找不到公共祖先, 要么 p或者 q就是公共祖先
- 进入递归, 从 root的左子树和右子树分别搜索公共祖先
- 如果左子树为空, 说明公共祖先一定在右子树, 直接返回右子树
- 如果右子树为空, 说明公共祖先一定在左子树, 直接返回左子树
- 如果两个子树都非空, 说明根结点就是公共祖先
- 如果两个子树都空, 这种情况就是 1中的 root为空
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
if (!root or root == p or root == q) return root;
TreeNode *left = lowestCommonAncestor(root -> left, p, q), *right = lowestCommonAncestor(root -> right, p, q);
if (!left) return right;
if (!right) return left;
return root;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q) return root;
TreeNode left = lowestCommonAncestor(root.left, p, q), right = lowestCommonAncestor(root.right, p, q);
if (left == null) return right;
if (right == null) return left;
return root;
}
}
Python:
# 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: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if root in (None, p, q):
return root
l, r = self.lowestCommonAncestor(root.left, p, q), self.lowestCommonAncestor(root.right, p, q)
return r if l is None else l if r is None else root
