572. Subtree of Another Tree 子树的判定

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.
给定两棵非空二叉树s和t，判定t是否为s的一棵子树。

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.

---

思路
递归思想，本节点不满足时则取两个子节点判定结果的逻辑或，本节点满足时则判定两个子节点结果的逻辑与。

/**
 * 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==nullptr) 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==nullptr && t==nullptr) return true;
        if(s==nullptr || t==nullptr) return false;
        if(s->val != t->val) return false;
        return isSame(s->left,t->left) && isSame(s->right,t->right);
    }
};