"Leetcode" Same Tree (2 solutions)

Same Tree

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.

Solution One: Recursion

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public:    BOOLIssametree (TreeNode *p, TreeNode *q) {if(!p &&!)q)return true; Else if(!p &&q)return false; Else if(P &&!)q)return false; Else        {            if(P->val! = q->val)return false; Else                returnIssametree (P->left, Q->left) && issametree (P->right, q->Right ); }    }};

Solution Two: Non-recursive

Set up two queues for hierarchy traversal, and check that the corresponding points are equal when entering the team

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public:    BOOLIssametree (TreeNode *p, TreeNode *q) {if(!p &&!)q)return true; Else if(!p &&q)return false; Else if(P &&!)q)return false; Else        {            if(P->val! = q->val)return false; Else{Queue<TreeNode*>LQ; Queue<TreeNode*>RQ;                Lq.push (P);                Rq.push (q);  while(!lq.empty () &&!Rq.empty ()) {TreeNode* Lfront =Lq.front (); TreeNode* Rfront =Rq.front ();                    Lq.pop ();                    Rq.pop (); if(!lfront->left &&!rfront->Left );//NULL                    Else if(!lfront->left && rfront->Left )return false; Else if(Lfront->left &&!rfront->Left )return false; Else                    {                        if(Lfront->left->val! = rfront->left->val)return false; Else{Lq.push (Lfront-Left ); Rq.push (Rfront-Left ); }                    }                    if(!lfront->right &&!rfront->Right );//NULL                    Else if(!lfront->right && rfront->Right )return false; Else if(Lfront->right &&!rfront->Right )return false; Else                    {                        if(Lfront->right->val! = rfront->right->val)return false; Else{Lq.push (Lfront-Right ); Rq.push (Rfront-Right ); }                    }                    }                return true; }        }    }};

"Leetcode" Same Tree (2 solutions)