Leetcode 0236: 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).”

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.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

The number of nodes in the tree is in the range [2, 10⁵].
-10⁹ <= Node.val <= 10⁹
All Node.val are unique.
p != q
p and q will exist in the tree.

Time complexity: O(n)
三种情况讨论：

1. 若当前根结点节点root 和二者之一相同 root == p 或者 root == q，则表示另外的点一定在root的左右子树其中一处，则最近的公共结点肯定是root。
2. p和q在root的同一侧，则他们最近公共祖先要么在root的左子树，要么在root的右子树，则直接递归到root.left和root.right进行搜索，若递归完后，左子树返回null表示没找到，那答案肯定是在右子树，同理，右子树返回null表示没找到，那答案肯定是在左子树
3. 若左右子树都找不到p和q的最近公共祖先，则表示p点和q点分别在不同的左右子树，则root就是他们的最近公共祖先

/**
 * 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);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if(left == null) return right;
        if(right == null) return left;
        return root;
    }
}