面试题68

题目描述：

测试用例：

思路：

• 利用二叉搜索树的性质，左子树都比根小，右子树都比根大，中序遍历为递增序列

基本流程：

1. 边界条件：

   1. 传入参数root为null，返回null

2. 循环查看root是否为空并处理以下三种情况：（利用二叉搜索树的性质）

   1. 都比root小，则两个节点都在root左侧，root赋值为左子树

   2. 都比root大，则两个节点都在root右侧，root赋值为右子树

   3. 一个比root小，另一个比root大，则为跳出循环条件，即找到了最近的公共祖先

代码：

import java.util.Objects;

public class P面试题68_IErChaSouSuoShuDeZuiJinGongGongZuXianLcof {

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

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

    //二叉搜索树中序遍历为递增，左孩子小于根小于右孩子
    //root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
    public static void main(String[] args) {
        Solution solution = new P面试题68_IErChaSouSuoShuDeZuiJinGongGongZuXianLcof().new Solution();
        // TO TEST
        TreeNode root = new TreeNode(6);
        TreeNode root1 = new TreeNode(2);
        TreeNode root2 = new TreeNode(8);
        TreeNode root3 = new TreeNode(0);
        TreeNode root4 = new TreeNode(4);
        TreeNode root5 = new TreeNode(7);
        TreeNode root6 = new TreeNode(9);
        TreeNode root7 = new TreeNode(3);
        TreeNode root8 = new TreeNode(5);


        root.left = root1;
        root.right = root2;
        root1.left = root3;
        root1.right = root4;
        root2.left = root5;
        root2.right = root6;
        root4.left = root7;
        root4.right = root8;

        root3.left = root3.right = root5.left = root5.right = root6.left = root6.right = root7.left = root7.right = root8.left = root8.right = null;

        System.out.println(solution.lowestCommonAncestor(root, root1, root4).val);

    }
    //leetcode submit region begin(Prohibit modification and deletion)

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     * int val;
     * TreeNode left;
     * TreeNode right;
     * TreeNode(int x) { val = x; }
     * }
     */
    /**
     *
     * 如果三种情况：
     *  1.都比根大：都在右侧
     *  2.都比根小：都在左侧
     *  3.一个比根大一个比根小 或者 存在等于根的：就是要找的祖先
     *
     *  还有一个意外情况为开始的根为null
     */
    class Solution {
        public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

            if (Objects.isNull(root))
                return null;

            while (Objects.nonNull(root)){

                if (root.val < p.val && root.val < q.val){
                    root = root.right;
                }else if (root.val > p.val && root.val > q.val){
                    root = root.left;
                }else{
                    break;
                }

            }

            return root;
        }
    }
}