剑指offer-面试题18：树的子结构

题目

public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
public TreeNode() {
// TODO Auto-generated constructor stub
}
}

解

//在树A中查找与树B根节点的值一样的结点，先根遍历
boolean HasSubTree(TreeNode pRoot1,TreeNode pRoot2){

boolean result=false;

if(pRoot1!=null && pRoot2!=null){
if(pRoot1.val==pRoot2.val)//找到树A中查找的根节点的值一样的结点
result=DoesTree1HaveTree2(pRoot1,pRoot2);
if(!result)
result=HasSubTree(pRoot1.left,pRoot2);//没找到或者
if(!result)
result=HasSubTree(pRoot1.right,pRoot2);
}

return result;
}

//判断树A中以R为根节点的子树是不是和树B具有相同的结构
boolean DoesTree1HaveTree2(TreeNode pRoot1, TreeNode pRoot2){

if(pRoot2==null)//树B的结点为空，循环终止
return true;
if(pRoot1==null)//树B的结点不为空，而树A对应的结点为空，循环终止
return false;
if(pRoot1.val!=pRoot2.val)//树Ａ和树Ｂ对应结点值不同，循环终止
return false;
return DoesTree1HaveTree2(pRoot1.left,pRoot2.left) && DoesTree1HaveTree2(pRoot1.right,pRoot2.right);
}