Subtree

Subtree

Description:

You have two every 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.

Notice

A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.

Example:

T2 is a subtree of T1 in the following case:

       1                  3
      / \                / 
T1 = 2   3        T2 =  4
    /
   4

T2 isn’t a subtree of T1 in the following case:

       1               3
      / \               \
T1 = 2   3        T2 =   4
    /
   4

分析：

判断T2是否是T1的子树，首先应该在T1中找到T2的根节点，且两棵子树必须完全相同。所以整个思路分为两步走：找根节点，判断两棵树是否全等；若不等，则需要判断是否是左/右子树的子树。

注意递归的先后顺序：需要先调用identical（调用.val前需要判断是否为null），再递归调用isSubtree判断左/右子树的情况。

identical的调用，时间复杂度近似O(n), 查根节点的时间复杂度随机平均为O(m), 故总的时间复杂度可近似为O(mn)。

代码如下：

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/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param T1, T2: The roots of binary tree.
     * @return: True if T2 is a subtree of T1, or false.
     */
    bool isSubtree(TreeNode *T1, TreeNode *T2) {
        // write your code here
        if (T2 == nullptr) return true;
        if (T1 == nullptr) return false;
        
        return identical(T1, T2) || isSubtree(T1->left, T2) || isSubtree(T1->right, T2);
    }
    
    bool identical(TreeNode *T1, TreeNode *T2) {
        if (T1 == nullptr && T2 == nullptr) return true;
        if (T1 == nullptr || T2 == nullptr) return false;
        if (T1->val != T2->val) return false;
        
        return identical(T1->left, T2->left) && identical(T1->right, T2->right);
    }
};