剑指offer-题18：树的子结构_dosetree1havetree2-优快云博客

本文介绍了一种算法，用于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功能，并提供了Java和Python两种语言的实现代码。

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题目描述

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）

实验平台：牛客网

解决思路：

这里写图片描述

java:

/**
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 = DoseTree1HavaTree2(root1, root2);
            }
            if (!result) {
                result = HasSubtree(root1.left, root2);
            }
            if (!result) {
                result = HasSubtree(root1.right, root2);
            }
        }
        return result;
    }

    public boolean DoseTree1HavaTree2(TreeNode root1, TreeNode root2) {
        if (root2 == null) {
            return true;
        }
        if (root1 == null) {
            return false;
        }
        if (root1.val != root2.val) {
            return false;
        }
        return DoseTree1HavaTree2(root1.left, root2.left) && DoseTree1HavaTree2(root1.right, root2.right);
    }
}

python:

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        result = False
        if pRoot1 is not None and pRoot2 is not None:
            if pRoot1.val == pRoot2.val:
                result = self.DoseTree1HaveTree2(pRoot1, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.left, pRoot2)
            if not result:
                result = self.HasSubtree(pRoot1.right, pRoot2)
        return result

    def DoseTree1HaveTree2(self, pRoot1, pRoot2):
        if pRoot2 is None:
            return True
        if pRoot1 is None:
            return False
        if pRoot2.val != pRoot1.val:
            return False
        return self.DoseTree1HaveTree2(pRoot1.right, pRoot2.right) and self.DoseTree1HaveTree2(pRoot1.left, pRoot2.left)