代码的鲁棒性：树的子结构

Posted 2020-08-04 SaraMorning

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

代码实现方式一：

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        if (root1 == null || root2 == null) {
            return false;
        }
        return isSubTree(root1, root2) || HasSubtree(root1.left, root2)
                || HasSubtree(root1.right, root2);
    }

    public boolean isSubTree(TreeNode root1, TreeNode root2) {
        if (root2 == null)
            return true;
        if (root1 == null)
            return false;
        if (root1.val == root2.val) {
            return isSubTree(root1.left, root2.left)
                    && isSubTree(root1.right, root2.right);
        } else
            return false;    }
}

 

代码实现方式二：

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        boolean result = false;
        if (root1 != null && root2 != null) {
            if (root1.val == root2.val) {
                result = isSubTree(root1, root2);
            }
            if (!result) {
                result = HasSubtree(root1.left, root2);
            }
            if (!result) {
                result = HasSubtree(root1.right, root2);
            }
        }
        return result;
    }

    public boolean isSubTree(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 != null)
            return false;
        if (root2 == null) {
            return true;
        }
        if (root1.val != root2.val) {
            return false;
        }
        return isSubTree(root1.left, root2.left)
                && isSubTree(root1.right, root2.right);
    }
}