[Leetcode] 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.

Idea one: The idea of pre-order recursion. Time complexity O (n), Spatial complexity O (LOGN)

　　

1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8  * };9  */Ten classSolution { One  Public: A     BOOLIssametree (TreeNode *p, TreeNode *q) { -         if(!p &&!q)return true; -         if(!p | |!q)return false; the          -         return(P->val = = q->val) && issametree (P->left, Q->left) && issametree (P->right, q->Right ); -     } -};

Idea two: The thought of hierarchical traversal. Time complexity O (n), Spatial complexity O (LOGN)

　　

1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8  * };9  */Ten classSolution { One  Public: A     BOOLIssametree (TreeNode *p, TreeNode *q) { -Queue<treenode *>s; - S.push (p); the S.push (q); -          -          while(!S.empty ()) { -p =S.front (); + S.pop (); -Q =S.front (); + S.pop (); A             if(!p &&!q)Continue; at             if(!p | |!q)return false; -             if(P->val! = q->val)return false; -              -S.push (p->Left ); -S.push (q->Left ); -S.push (p->Right ); inS.push (q->Right ); -         } to          +         return true; -     } the};

[Leetcode] Same Tree