LeetCode #572 - Subtree of Another Tree

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本文探讨了二叉树中子树匹配的问题，并提供了一种解决方案。通过定义两个递归函数isSubtree和isSubtree_from_cur来实现对两棵树的结构和节点值的比较。

题目描述：

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

Given tree t:

   4 
  / \
 1   2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

Given tree t:

   4
  / \
 1   2

Return false.

给定两个二叉树t和s，判断t是否为s的子树。一般的方法是比较根结点，根据比较结果进行递归，由于不一定从根结点开始匹配，所以定义两个递归函数isSubtree和isSubtree_from_cur，如果确定从根结点开始匹配的话，那么左子树和右子树都要从对应的根结点开始匹配。

class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if(t==NULL&&s==NULL) return true;
        else if(t==NULL&&s!=NULL) return false;
        else if(t!=NULL&&s==NULL) return false;
        if(s->val==t->val) 
            return isSubtree(s->left,t)||isSubtree(s->right,t)||
                   (isSubtree_from_cur(s->left,t->left)&&isSubtree_from_cur(s->right,t->right));
        else return isSubtree(s->left,t)||isSubtree(s->right,t);
    }
    
    bool isSubtree_from_cur(TreeNode* s, TreeNode* t){
        if(t==NULL&&s==NULL) return true;
        else if(t==NULL&&s!=NULL) return false;
        else if(t!=NULL&&s==NULL) return false;
        if(t->val!=s->val) return false;
        return isSubtree_from_cur(s->left,t->left)&&isSubtree_from_cur(s->right,t->right);
    }
};