网易雷火笔试-编程题（1）

题目描述：

给定一个数组tree和两个节点p,q,计算最低公共祖先的值。

输入：一个数组和两个节点值

输出：计算最低公共祖先的值

如：

输入：

1,2,3,4,5,6,7，-1，-1,8,9

4

3

输出：1

思路：

最低公共祖先的题目肯定会求，本题麻烦就在于要把输入的数组转换成二叉树，另外两个值转换成二叉树中的节点值。

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

class TreeNode{
	int val;
	TreeNode left;
	TreeNode right;
	public TreeNode(int x) {
		val = x;
		left = null;
		right = null;
	}
}
public class Main3 {
	static TreeNode t1;
	static TreeNode t2;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine();
		int p = sc.nextInt();
		int q = sc.nextInt();
		String[] strs = s.split(",");
		int[] tree = new int[strs.length];
		for(int i = 0; i < strs.length; i ++) {
			tree[i] = Integer.parseInt(strs[i]);
		}
		TreeNode root = buildTree(tree, p, q);
		TreeNode res =  lowestCommonAncestor(root, t1, t2);
		System.out.println(res.val);
	}
	public static TreeNode buildTree(int[] nums, int x, int y) {
		if(nums == null || nums.length == 0) return null;
		Queue q = new LinkedList<>();
		TreeNode root = new TreeNode(nums[0]);
		q.offer(root);
		int i = 1;
		if(root.val == x) t1 = root;
		if(root.val == y) t2 = root;
		while(i < nums.length) {
			int size = q.size();
			while(size -- > 0 && i < nums.length) {
				TreeNode node = q.poll();
				node.left = new TreeNode(nums[i++]);
				node.right = new TreeNode(nums[i++]);
				q.offer(node.left);
				q.offer(node.right);
				if(node.left.val == x) t1 = node.left;
				if(node.left.val == y) t2 = node.left;
				if(node.right.val == x) t1 = node.right;
				if(node.right.val == y) t2 = node.right;
			}
		}
		return root;
	}
	public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root.val  == -1) return null;
		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 || left.val == -1){
            return right;
        }else if(right == null || right.val == -1){
            return left;
        }else{
            return root;
        }
    }
}