DFS:100. Same Tree

比较根节点，然后比较左子树，然后是右子树。

如果p为空且q为空则true，如果p不为空且q不为空，比较pq的值和左子树右子树，其他情况都是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 isSameTree(TreeNode* p, TreeNode* q) {        if(p == NULL && q == NULL)            return true;        if(p != NULL && q != NULL)            return p->val == q->val && isSameTree(p->left,q->left) && isSameTree(p->right, q->right);        else            return false;    }};

再少几行：

bool isSameTree(TreeNode *p, TreeNode *q) {    if (p == NULL || q == NULL) return (p == q);    return (p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right));}