The substructure of a tree (the sword refers to an offer) recursive __

Child structure of the tree number of participants: 1611 time limit: 1 seconds space limit: 32768K pass ratio: 18.19% best record: 0 ms|0k (from Mearo) Topic DescriptionEnter two binary tree a,b to determine whether B is a sub structure.

Link: http://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?rp=1&ru=/ta/coding-interviews&qru =/ta/coding-interviews/question-ranking

Train of thought: To determine whether a B, then, first find B's root node in a position, and then start from the node, compare the left and right subtree is equal;

I am not very good at the tree problem, originally want to own a achievements, the result, incredibly wood has output ... Helpless Ah.

My idea is to build the first number of the team, and then pick up, into the stack, ... Each stack node from the team to take the left and right node into the stack, with a linked list;

By the way, after finding the root node of B in a, a recursive decision to exit is a condition where B arrives null, returns True, or the value of a node and the value of the B node are unequal + or a ratio b is first to NULL, returning false, otherwise recursive to the left and right nodes of the node;

/* struct TreeNode {int val;
	struct TreeNode *left;
	struct TreeNode *right; TreeNode (int x): Val (x), left (null), right (null) {}};*/class Solution {public:bool hassubtree (treenode* Proo
        T1, treenode* pRoot2) {bool Bo=false;
            if (pRoot1 && pRoot2) {if (proot1->val==proot2->val) Bo=judgetree (PROOT1,PROOT2);
            if (!bo) Bo=hassubtree (PROOT1->LEFT,PROOT2);
        if (!bo) Bo=hassubtree (PROOT1->RIGHT,PROOT2);
    return Bo;
        BOOL Judgetree (TreeNode *p1,treenode *p2) {if (!P2) return true;
        if (!P1) return false;
        if (P1->val!=p2->val) return false;
    Return Judgetree (P1->left,p2->left) && judgetree (p1->right,p2->right); }

};