剑指offer26 树的子结构

剑指offer26 树的子结构

    输入两棵二叉树A，B，判断B是不是A的子结构。

    我们规定空树不是任何树的子结构。

样例

树A：

         8
        / \
       8   7
      / \
     9   2
        / \
       4   7

树B：

       8
      / \
     9   2

返回 true ,因为B是A的子结构。

思路：

    第一步，遍历tree1中所有与tree2头节点相等的结点
    第二步，找到相同的头节点，然后依次判断其左子树与右子树是否相等。

AcWing-37 c++ code:

/**
 * 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 hasSubtreeCore(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2 == NULL){
            return true;
        }
        if(pRoot1 == NULL){
            return false;
        }
        if(pRoot1->val != pRoot2->val){
            return false;
        }
        
        return hasSubtreeCore(pRoot1->left, pRoot2->left) && hasSubtreeCore(pRoot1->right, pRoot2->right);

    }
    bool hasSubtree(TreeNode* pRoot1, TreeNode* pRoot2){
        bool result = false;
        if(pRoot1 && pRoot2){
            if(pRoot1->val == pRoot2->val){
                result = hasSubtreeCore(pRoot1, pRoot2);
            }
            if(!result){
                result = hasSubtree(pRoot1->left, pRoot2);
            }
            if(!result){
                result = hasSubtree(pRoot1->right, pRoot2);
            }
        }
        return result;
    }
    
};

AcWing-37 python code:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def hasSubtreeCore(self, pRoot1, pRoot2):
        if pRoot2 == None:
            return True
        if pRoot1 == None:
            return False
        if pRoot1.val != pRoot2.val:
            return False
        return self.hasSubtreeCore(pRoot1.left, pRoot2.left) and self.hasSubtreeCore(pRoot1.right, pRoot2.right)
        
    def hasSubtree(self, pRoot1, pRoot2):
        """
        :type pRoot1: TreeNode
        :type pRoot2: TreeNode
        :rtype: bool
        """
        result = False
        if pRoot1 and pRoot2:
            if pRoot1.val == pRoot2.val:
                result = self.hasSubtreeCore(pRoot1, pRoot2)
            if not result:
                result = self.hasSubtree(pRoot1.left, pRoot2)
            if not result:
                result = self.hasSubtree(pRoot1.right, pRoot2)
        return result