LowestCommonAncestorOfABinaryTree.java

/**
 * <p>给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。</p>
 *
 * <p><a href="https://baike.baidu.com/item/%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88/8918834?fr=aladdin" target="_blank">百度百科</a>中最近公共祖先的定义为：“对于有根树 T 的两个节点 p、q，最近公共祖先表示为一个节点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（<strong>一个节点也可以是它自己的祖先</strong>）。”</p>
 *
 * <p> </p>
 *
 * <p><strong>示例 1：</strong></p>
 * <img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
 * <pre>
 * <strong>输入：</strong>root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
 * <strong>输出：</strong>3
 * <strong>解释：</strong>节点 <code>5 </code>和节点 <code>1 </code>的最近公共祖先是节点 <code>3 。</code>
 * </pre>
 *
 * <p><strong>示例 2：</strong></p>
 * <img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
 * <pre>
 * <strong>输入：</strong>root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
 * <strong>输出：</strong>5
 * <strong>解释：</strong>节点 <code>5 </code>和节点 <code>4 </code>的最近公共祖先是节点 <code>5 。</code>因为根据定义最近公共祖先节点可以为节点本身。
 * </pre>
 *
 * <p><strong>示例 3：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>root = [1,2], p = 1, q = 2
 * <strong>输出：</strong>1
 * </pre>
 *
 * <p> </p>
 *
 * <p><strong>提示：</strong></p>
 *
 * <ul>
 * <li>树中节点数目在范围 <code>[2, 10<sup>5</sup>]</code> 内。</li>
 * <li><code>-10<sup>9</sup> <= Node.val <= 10<sup>9</sup></code></li>
 * <li>所有 <code>Node.val</code> <code>互不相同</code> 。</li>
 * <li><code>p != q</code></li>
 * <li><code>p</code> 和 <code>q</code> 均存在于给定的二叉树中。</li>
 * </ul>
 * <div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>二叉树</li></div></div><br><div><li>👍 1398</li><li>👎 0</li></div>
 */

package leetcode3;

public class LowestCommonAncestorOfABinaryTree {

    public static void main(String[] args) {
        Solution solution = new LowestCommonAncestorOfABinaryTree().new Solution();
    }

    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }

    /**
     * 可参考这篇题解：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/solution/236-er-cha-shu-de-zui-jin-gong-gong-zu-xian-hou-xu/
     * DFS
     */
    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;
        }
    }
}