【剑指 Offer 26. 树的子结构】-优快云博客

本文介绍了一种有效的算法来判断一棵二叉树是否为另一棵二叉树的子结构。通过递归方法实现了该算法，并给出了正确的实现代码。

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【解题思路】

我用了以下的方法，发现在测试样例：A = [1,0,1,-4,-3]，B = [1,-4]，会出现报错，因为这种方法不能判断树不连续地相等。因此，必须使用一个子函数，来连续地判断B是不是A的子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if(A != null && B != null)
        {
            if(A.val == B.val)
            {
                if(B.left != null && B.right != null)
                {
                    return (isSubStructure(A.left, B.left) && isSubStructure(A.right, B.right)) || isSubStructure(A.left, B) || isSubStructure(A.right, B);
                }
                else if(B.left != null)
                {
                    return isSubStructure(A.left, B.left);
                }
                else if(B.right != null)
                {
                    return isSubStructure(A.right, B.right);  
                }
                return true;
            }
            return (isSubStructure(A.left, B) || isSubStructure(A.right, B));
        }
        else if(A != null && B == null)
        {
            return false;
        }
        else if(A == null && B != null)
        {
            return false;
        }
        else
        {
            return true;
        }
    }
}

以下为正确代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        return (A != null && B!= null )  && (isSubTree(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B));
    }

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