Q4.8 To decide if T2 is a subtree of T1

Q:You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1

A:DFS.

T1当前节点是否与T2的根节点相等。

如果不相等，查看T2是不是T1的左、右子树的子树

如果相等，查看T1的左右子树和T2的左右子树是否一致

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() {}
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

//初始化数据 
TreeNode *dfs(vector<int> &num, int start, int end) {
    if (start == end) {
        return NULL;
    }
    int mid = start + (end-start)/2;
    TreeNode *root = new TreeNode(num[mid]);
    root->left = dfs(num, start, mid);
    root->right = dfs(num, mid+1, end);
    return root;
}

TreeNode *sortedArrayToBST(vector<int> &num) {
    const int n = num.size();
    if (n == 0) {
        return NULL;
    }
    return dfs(num,0,n);
}
    
//算法实现 
bool matchTree(TreeNode *p, TreeNode *q) {
	if (!p && !q) {
		return true;
	}
	if (!p || !q) {
		return false;
	}
	if (p->val != q->val) {
		return false;
	} 
	return matchTree(p->left, q->left) && matchTree(p->right, q->right);
	
}

bool subTree(TreeNode *root1, TreeNode *root2) {
	if (!root1) {
		return false;
	}
	if (root1->val == root2->val) {
		return matchTree(root1, root2);
	}
	return matchTree(root1->left, root2)||matchTree(root1->right, root2);
}

bool containsTree(TreeNode *root1, TreeNode *root2) {
	if (!root2) {
		return true;
	}
	return subTree(root1, root2);
}


vector<int> inorderTraversal(TreeNode *root) {  
    vector<int> res;  
    if (!root) {  
        return res;  
    }  
    TreeNode *cur = root;  
    while (cur) {  
        if (cur->left == NULL) {  
            res.push_back(cur->val);  
            cur = cur->right;  
        } else {  
            TreeNode *node = cur->left;  
            while (node->right && node->right != cur) {  
                node = node->right;  
            }  
            if (!node->right) {  
                node->right = cur;  
                cur = cur->left;  
            } else {  
                res.push_back(cur->val);  
                node->right = NULL;  
                cur = cur->right;  
            }  
        }  
    }  
      
      
    return res;  

    }  

int main() {
	vector<int> a;
	for (int i = 1; i < 10; i++) {
		a.push_back(i);
	}
	TreeNode *root1 = sortedArrayToBST(a);
	a.clear();
	a = inorderTraversal(root1);
	for (int i = 0; i < a.size(); i++) {
		cout<<a[i]<<" ";
	}
	cout<<endl;
	vector<int> b;
	for (int i = 1; i < 5; i++) {
		b.push_back(i);
	}
	TreeNode *root2 = sortedArrayToBST(b);
	b.clear();
	b = inorderTraversal(root2);
	for (int i = 0; i < b.size(); i++) {
		cout<<b[i]<<" ";
	}
	cout<<endl;
	cout<<containsTree(root1, root2)<<endl;
	return 0;
}