【重点】【二叉树】【递归】剑指offer——面试题18：树的子结构

本文链接：https://blog.csdn.net/Allenlzcoder/article/details/79588292

本文介绍三种不同的算法实现方式来判断一棵二叉树是否为另一棵二叉树的子结构，包括直接递归、先查找再判断以及优化的一遍递归一遍判断的方法。

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递归

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubStructure(self, A: Optional[TreeNode], B: Optional[TreeNode]) -> bool:
        if not A or not B: # 若A与B其中一个为空,立即返回False
            return False
        return self.isSame(A, B) or self.isSubStructure(A.left, B) or self.isSubStructure(A.right, B)
    
    def isSame(self, A, B):
        if not B:   # 若B走完了,说明查找完毕,B为A的子结构
            return True
        elif not A: # 若B不为空并且A为空或者A与B的值不相等,直接可以判断B不是A的子结构
            return False
        elif A.val != B.val:
            return False
        # 当A与B当前节点值相等,若要判断B为A的子结构
        # 还需要判断B的左子树是否为A左子树的子结构 && B的右子树是否为A右子树的子结构
        # 若两者都满足就说明B是A的子结构,并且该子结构以A根节点为起点
        return self.isSame(A.left, B.left) and self.isSame(A.right, B.right)

Java

class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if (A == null || B == null) {
            return false;
        }

        return isSame(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B);
    }

    public boolean isSame(TreeNode a, TreeNode b) {
        if (b == null) {
            return true;
        } else if (a == null) {
            return false;
        }else if (a.val != b.val) {
            return false;
        }

        return isSame(a.left, b.left) && isSame(a.right, b.right);
    }
}

// 先把可能的根节点找出来再去判断，速度慢！
class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if (Objects.isNull(B)) {
            return false;
        }
        List<TreeNode> beginNodeList = new ArrayList<>();
        findBeginNode(A, B.val, beginNodeList);
        for (TreeNode beginNode : beginNodeList) {
            if (isSameTree(beginNode, B)) {
                return true;
            }
        }

        return false;
    }

    public void findBeginNode(TreeNode A, int val, List<TreeNode> beginNodeList) {
        if (Objects.isNull(A)) {
            return;
        } else if (A.val == val) {
            beginNodeList.add(A);
        }
        findBeginNode(A.left, val, beginNodeList);
        findBeginNode(A.right, val, beginNodeList);
    }

    public boolean isSameTree(TreeNode a, TreeNode b) {
        if (Objects.isNull(b)) {
            return true;
        } else if (Objects.isNull(a) && !Objects.isNull(b)) {
            return false;
        } else if (a.val != b.val) {
            return false;
        } else {
            boolean left = isSameTree(a.left, b.left);
            boolean right = isSameTree(a.right, b.right);
            return left && right;
        }
    }
}

答案2

一遍递归，一遍判断，速度略快！！！

/**
 * 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) {
        boolean res = false;
        if (!Objects.isNull(A) && !Objects.isNull(B)) {
            if (A.val == B.val) {
                res = isSameTree(A, B);
            }
            if (!res) {
                res = isSubStructure(A.left, B);
            }
            if (!res) {
                res = isSubStructure(A.right, B);
            }
        }

        return res;
    }

    public boolean isSameTree(TreeNode a, TreeNode b) {
        if (Objects.isNull(b)) {
            return true;
        } else if (Objects.isNull(a) && !Objects.isNull(b)) {
            return false;
        } else if (a.val != b.val) {
            return false;
        } else {
            boolean left = isSameTree(a.left, b.left);
            boolean right = isSameTree(a.right, b.right);
            return left && right;
        }
    }
}

Solution1：
此题答案是抄的书上的，要记忆并熟练运用关于二叉树的递归思想！！！

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2){
        bool result = false;
        //先找到相同的根结点的值
        if(pRoot1 != NULL && pRoot2 != NULL){
            if(pRoot1->val == pRoot2->val)
                result = DoesTree1HaveTree2(pRoot1, pRoot2);
            if(!result)
                result = HasSubtree(pRoot1->left, pRoot2);
            if(!result)
                result = HasSubtree(pRoot1->right, pRoot2);
        }
        return result;
    }
    //判断子树结构是否一致
    bool DoesTree1HaveTree2(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2 == NULL) //已经到达树2的根结点，返回true
            return true;
        if(pRoot1 == NULL) //已经到了树1的根结点，但还未到达树2的根结点，返回false
            return false;
        if(pRoot1->val != pRoot2->val) //值不相同，则返回false
            return false;
        //递归判断
        return DoesTree1HaveTree2(pRoot1->left, pRoot2->left) && DoesTree1HaveTree2(pRoot1->right, pRoot2->right);
    }
};