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

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