剑指offer第17题（树的子结构）

题目：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）

思想：现根据在树A中查找与根节点的值一样的节点R。紧接着判断树A中以节点R为根节点的子树和树b是不是具有相同的结构。

java代码：

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

python代码：

class Solution:    def HasSubtree(self, pRoot1, pRoot2):        # write code here        result = False        if pRoot1 and pRoot2:            if pRoot1.val == pRoot2.val:                result = self.IsSubtree(pRoot1, pRoot2)            if not result:                result = self.HasSubtree(pRoot1.left, pRoot2)            if not result:                result = self.HasSubtree(pRoot1.right, pRoot2)        return result    def IsSubtree(self, pRoot1, pRoot2):        if not pRoot2:            return True        if not pRoot1:            return False        if pRoot1.val != pRoot2.val:            return False        return self.IsSubtree(pRoot1.left, pRoot2.left) and self.IsSubtree(pRoot1.right, pRoot2.right)