本文主要是介绍算法：最小公共祖先236. Lowest Common Ancestor of a Binary Tree，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

236. 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).”

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.

解法：回溯递归判断左右子树是否为空

回溯判断，

1. 如果左右节点都不为空，那么返回root
2. 如果左右节点都为null，那么返回null
3. 如果左节点不为空，那么返回返回左节点，否则返回右结点。

下面的代码，隐藏了第2个逻辑，因为如果都为空，返回left和right都一样。

/**
 * 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 && right != null) return root;
        return left != null ? left : right;
    }
}

参考

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/discuss/65225/4-lines-C%2B%2BJavaPythonRuby

这篇关于算法：最小公共祖先236. Lowest Common Ancestor of a Binary Tree的文章就介绍到这儿，希望我们推荐的文章对大家有所帮助，也希望大家多多支持为之网！