# "Leetcode from zero single brush" Same Tree

Source: Internet
Author: User

Yes, I'm Wu. 2009 fans, from today, imitate the "from zero single row" series, Vegetable chicken single brush leetcode!

Topic:

Given binary trees, write a function to check if they is equal or not. The binary trees is considered equal if they is structurally identical and the nodes has the same value.

Verify that the two trees are equal. Using the idea of recursion , if the Val of the tree nodes are equal, then go to each of the trees to calculate whether the subtree is equal.

Note the recursive endpoint: the output is true when all two trees are empty trees . So I got this:

`/** * Definition for binary tree * struct TreeNode {*     int val; *     TreeNode *left; *     TreeNode *right; *     Tre Enode (int x): Val (x), left (null), right (NULL) {} *}; */class Solution {public:    bool Issametree (TreeNode *p, TreeNode *q) {        if (p = = NULL && q = = null)            retur n true;        else if (P->val = = Q->val)            return (Issametree (P->left, Q->left) && issametree (P->right, q >right));        else             return false;    }};`
Well, not through ... When Enter an empty tree, and when compared to a non-empty tree, the empty tree simply cannot fetch the Val value。 So to be the case of entering an empty tree first (if it is placed in a later branch, or if an empty tree evaluation is encountered) excludes all。 Get:

`/** * Definition for binary tree * struct TreeNode {*     int val; *     TreeNode *left; *     TreeNode *right; *     Tre Enode (int x): Val (x), left (null), right (NULL) {} *}; */class Solution {public:    bool Issametree (TreeNode *p, TreeNode *q) {        if (p = = NULL && q = = null)            retur n true;        else if ((p==null && q!=null) | | (q==null && p!=null) | | (P->val! = q->val))            return false;           else if (P->val = = Q->val)            return (Issametree (P->left, Q->left) && issametree (P->right, q >right));}    };`

"Leetcode from zero single brush" Same Tree

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