面试题18：树的子结构

package com.hb.jzoffer;

import javax.swing.text.Position.Bias;

import offer.utilities.BinaryTreeNode;
//面试题18：树的子结构
/*
 * 题目：输入两颗二叉树 A 和 B ，判断 B 是不是 A 的子结构。
 */
public class SubstructureInTree_18 {
    
    public boolean  hasSubTree(BinaryTreeNode pRoot1 , BinaryTreeNode  pRoot2){
        boolean result = false;
        
        if(pRoot1 != null  && pRoot2 != null){
            if(pRoot1.val == pRoot2.val){
                result = doesTree1HavaTree2(pRoot1 , pRoot2);
            }
            if(!result){
                result = hasSubTree(pRoot1.pLeft, pRoot2);
            }
            if(!result){
                result = hasSubTree(pRoot1.pRight, pRoot2);
            }
        }
        
        return  result ;
    }

    public boolean doesTree1HavaTree2(BinaryTreeNode pRoot1, BinaryTreeNode pRoot2) {
        // TODO Auto-generated method stub
        if(pRoot2 == null){
            return true ;
        }
        
        if(pRoot1 == null){
            return false;
        }
        
        if(pRoot1.val != pRoot2.val){
            return false;
        }
        
        return doesTree1HavaTree2(pRoot1.pLeft , pRoot2.pLeft) && doesTree1HavaTree2(pRoot1.pRight, pRoot2.pRight);
    }
    
    public static void main(String[] args) {
        BinaryTreeNode  rootA = new BinaryTreeNode(8) ;
        BinaryTreeNode  nodeA_1 = new BinaryTreeNode(8);
        BinaryTreeNode  nodeA_2= new BinaryTreeNode(7);
        
        BinaryTreeNode  nodeA_1_1 = new BinaryTreeNode(9);
        BinaryTreeNode  nodeA_1_2 = new BinaryTreeNode(2);
        
        BinaryTreeNode  nodeA_1_2_1 = new BinaryTreeNode(4);
        BinaryTreeNode  nodeA_1_2_2 = new BinaryTreeNode(7);
        
        rootA.pLeft = nodeA_1 ;
        rootA.pRight = nodeA_2;
        nodeA_1.pLeft = nodeA_1_1;
        nodeA_1.pRight = nodeA_1_2;
        
        nodeA_1_2.pLeft = nodeA_1_2_1;
        nodeA_1_2.pRight = nodeA_1_2_2;
        
        
        BinaryTreeNode  rootB = new BinaryTreeNode(8) ;
        BinaryTreeNode  nodeB_1 = new BinaryTreeNode(9) ;
        BinaryTreeNode  rootB_2 = new BinaryTreeNode(2) ;
        rootB.pLeft = nodeB_1;
        rootB.pRight = rootB_2;
        
        SubstructureInTree_18  subtree = new SubstructureInTree_18();
        boolean  result = subtree.hasSubTree(rootA, rootB);
        System.out.println(result);
        
    }
    
    

}