判断一个节点是不是另一个的节点的子节点；

最新推荐文章于 2023-12-15 11:17:20 发布

原创最新推荐文章于 2023-12-15 11:17:20 发布 · 1.6k 阅读

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#tree #subnode

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本文详细介绍了如何使用Java实现一个简单的二叉树结构，并通过递归方法判断一个节点是否为另一个节点的子节点。代码示例清晰易懂，适合初学者理解和实践。

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

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

public class Test {

    public static void main(String[] args) {

        //测试用节点数组
        TreeNode[] tree = { new TreeNode(6), new TreeNode(2), new TreeNode(8),
                new TreeNode(0), new TreeNode(4), new TreeNode(7),
                new TreeNode(9), new TreeNode(3), new TreeNode(5) };
        tree[0].left = tree[1];
        tree[0].right = tree[2];
        tree[1].left = tree[3];
        tree[1].right = tree[4];
        tree[2].left = tree[5];
        tree[2].right = tree[6];
        tree[4].left = tree[7];
        tree[4].right = tree[8];

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) 
            {
                System.out.println("higher: t" + i + " ,lower: t" + j + " : "+ isSubNode(tree[i],tree[j]));
            }
        }
    }

    private static boolean isSubNode(TreeNode higher, TreeNode lower) {

        if (higher.left == null && higher.right == null)
            return false;
        if (higher.left == lower || higher.right == lower)
            return true;
        TreeNode tempLeft = higher.left;
        TreeNode tempRight = higher.right;
        return isSubNode(tempLeft, lower) || isSubNode(tempRight, lower);

    }
}