菜鸡刷Leetcode之236. Lowest Common Ancestor of a Binary Tree Medium

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

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.

分析：

首先，树的问题，往往涉及递归，之所以涉及递归是因为每一棵树都是由相同结构的多个子树构成。本题可以利用递归的方式，相当于从叶子节点开始，递归地返回如下信息：当前树（或子树）同时包含p和q，则函数返回的结果就是它们的最低公共祖先。如果它们其中之一在子树当中，则结果是它们中的一个。如果它们都不在子树里，则结果返回NULL。

class Solution {
public:
    TreeNode * lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        // 虽然下面两个if可以写到一起，但所表达的意思是不一样的
        if (root == nullptr) {
            return nullptr;
        }
        if (p == root || q == root) {
            return root;
        }
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if (left == nullptr) {  // 如果有一个分支为空，另一个不为空，则返回不为空的节点
            return right;
        }
        else if (right == nullptr) {
            return left;
        }
        else  // 如果两个都不为空，则返回二者当前最低祖先节点
            return root;
    }
};

总结：

书上对这道题的做法麻烦了。首先书上并不是二叉树，而是一颗普通树，所以next存在一个vector里。
其次，书上的做法是先找到p和q两条路径，从根节点出发到p和q分别构成两条链表；
最后，寻找这两条链表的第一个分叉的地方。

注：

如果本题的树是二叉搜索树或者有指向父节点的指针，则思路能更加简单。

菜鸡刷Leetcode之236. Lowest Common Ancestor of a Binary Tree Medium

原文：https://www.cnblogs.com/Flash-ylf/p/11494780.html