572. Subtree of Another Tree

最新推荐文章于 2024-08-31 10:23:18 发布

无名_1989

最新推荐文章于 2024-08-31 10:23:18 发布

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本文介绍了一种算法，用于检查一棵二叉树是否为另一棵二叉树的子结构。通过递归方法验证节点值及结构的一致性，并利用栈进行遍历，实现了高效的子树匹配。

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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.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if(!s) return false;
        if(isSame(s,t)) return true;
        return isSubtree(s->left, t) || isSubtree(s->right, t);
    }
    
    bool isSame(TreeNode *s , TreeNode *t)
    {
        if(!s&&!t) return true;
        if(!s||!t) return false;
        if(s->val != t->val) return false;
        return isSame(s->left, t->left) && isSame(s->right, t->right);
    }
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    bool isST(TreeNode *n, TreeNode *t)
    {
        if(n == NULL && t == NULL) return true;
        if(n == NULL && t!= NULL || n != NULL && t == NULL || n->val != t->val ) return false;
        bool bl = isST(n->left, t->left);
        bool br = isST(n->right, t->right);
        return bl && br;
    }
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        stack<TreeNode *> st;
        st.push(s);
        while(!st.empty())
        {
            TreeNode * node = st.top(); st.pop();
            if(node->left) st.push(node->left);
            if(node->right) st.push(node->right);
            if(node->val == t->val && isST(node, t)) return true;
        }
        return false;
    }
};