• [LeetCode] Subtree of Another Tree


    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 scould also be considered as a subtree of itself.

    Example 1:
    Given tree s:

         3
        / 
       4   5
      / 
     1   2
    

    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:

         3
        / 
       4   5
      / 
     1   2
        /
       0
    

    Given tree t:

       4
      / 
     1   2
    

    Return false.

    判断t是否为s的子树。首先这是一个判断一棵数的子树是否与另一棵树相同,则使用一个判断两树是否相同的函数,然后将s的每一个节点的子树与t比较,相同则返回true,否则返回false。

    class Solution {
    public:
        bool isSubtree(TreeNode* s, TreeNode* t) {
            if (s == nullptr)
                return false;
            if (isSameTree(s, t))
                return true;
            return isSubtree(s->left, t) || isSubtree(s->right, t);
        }
        bool isSameTree(TreeNode* s, TreeNode* t) {
            if (s == nullptr && t == nullptr)
                return true;
            if (s == nullptr || t == nullptr)
                return false;
            if (s->val == t->val)
                return isSameTree(s->left, t->left) && isSameTree(s->right, t->right);
            return false;
        }
    };
    // 25 ms

    先对两棵树进行先序遍历,注意的是先序遍历的时候需要在节点值前加入’#‘以区分节点值,并注意区分叶节点的左”lnull“右”rnull“空指针。并将先序遍历的结果放在一个字符串中,然后对两个字符串进行模式匹配。由于文本串和模式串都比较小,所以使用简单的基于蛮力算法的模式匹配算法。

    class Solution {
    public:
        bool isSubtree(TreeNode* s, TreeNode* t) {
            string tree1 = preOrder(s, true);
            string tree2 = preOrder(t, true);
            return match(tree1, tree2);
        }
        string preOrder(TreeNode* root, bool left) {
            if (root == nullptr) {
                if (left)
                    return "lnull";
                else
                    return "rnull";
            }
            return "#" + to_string(root->val) + " " + preOrder(root->left, true) + " " + preOrder(root->right, false);
        }
        bool match(string s1, string s2) {
            int n = s1.size(), m = s2.size();
            int i = 0, j = 0;
            while (j < m && i < n) {
                if (s2[j] == s1[i]) {
                    i++;
                    j++;
                }
                else {
                    i -= (j - 1);
                    j = 0;
                }
            }
            if (j == m)
                return true;
            else
                return false;
        }
    };
    // 42 ms
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  • 原文地址:https://www.cnblogs.com/immjc/p/7218763.html
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