JZ-076 - Lowest common ancestor of two nodes in tree

The lowest common ancestor of two nodes in the tree

Topic description

binary search tree

 * 二叉查找树中，两个节点 p, q 的公共祖先 root 满足 root.val >= p.val && root.val <= q.val。

topic link : [the lowest common ancestor of two nodes in the tree]()

code

/**
 * 标题：树中两个节点的最低公共祖先
 */
public class Jz76 {

    /**
     * 二叉查找树
     * 二叉查找树中，两个节点 p, q 的公共祖先 root 满足 root.val >= p.val && root.val <= q.val。
     *
     * @param root
     * @param p
     * @param q
     * @return
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return root;
        }
        if (root.val > p.val && root.val > q.val) {
            return lowestCommonAncestor(root.left, p, q);
        }
        if (root.val < p.val && root.val < q.val) {
            return lowestCommonAncestor(root.right, p, q);
        }
        return root;
    }

    /**
     * 普通二叉树
     * 在左右子树中查找是否存在 p 或者 q，如果 p 和 q 分别在两个子树中，那么就说明根节点就是最低公共祖先。
     *
     * @param root
     * @param p
     * @param q
     * @return
     */
    public TreeNode lowestCommonAncestor2(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);
        return left == null ? right : right == null ? left : root;
    }

    public static void main(String[] args) {

    }
}

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