LeetCode 236. Lowest Common Ancestor of a Binary Tree (二叉树的最小公共祖先)
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]
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
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.
Reference Answer
思路分析
注意是BST，那么使用分而治之的策略，用递归来找到左边和右边的最低的公共祖先。
最低公共祖先的定义是，在一个二叉树中，我们能找到的最靠近叶子的节点，该节点同时是p和q的祖先节点。注意，如果p或者q本身也可以作为自己的祖先。
如果用递归的话，最重要的还是明白递归函数的作用是什么。这个题里面lowestCommonAncestor(root, p, q)函数的作用是判断p和q在root树中最低的公共祖先是什么，返回值是公共祖先。
这个题的模式叫做devide and conquer. 如果当前节点等于其中的p和q某一个节点，那么找到了节点，返回该节点，否则在左右子树分别寻找。
左右子树两个返回的是什么呢？按照该递归函数的定义，即找到了左子树和右子树里p和q的公共祖先，注意祖先可以是节点自己。然后根据左右侧找到的节点做进一步的判断。
如果左右侧查找的结果都不为空，说明分别找到了p和q，那么LCA就是当前节点。否则就在不为空的那个结果就是所求。
自己想着直接遍历输出所有可能路径，再依据p，q在路径的索引值判断其层数先后顺序，最后再依据上述条件输出，但结果一直出错，暂且先放一波，个人觉着这个思路是没问题的。
Code
# 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 not root or root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
return left if left else right
补充一个 C++ 版：
/**
* 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 == NULL || p == root || q == root){
return root;
}
TreeNode* left = lowestCommonAncestor(root->left, p, q);
TreeNode* right = lowestCommonAncestor(root->right, p, q);
if(left != NULL && right != NULL){
return root;
}
return left ? left : right;
}
};
Note
- 这道题是典型的DFS(深度优先遍历)，之前大都是使用BFS（广度优先遍历），这种深度优先遍历思想跟自己想的输出所有可能路径近似，但这种做法无意简单到爆，注意学习其解题思路。
参考文献
[1] https://blog.csdn.net/fuxuemingzhu/article/details/51290289