[剑指Offer]26-树的子结构

题意

判断一棵树(参数二)是不是另一棵树(参数一)的子结构。

题解

递归第一棵树，找两棵树中值一样的节点。若找到后，用另一个函数判断以相同值得节点为根的树2是不是树1的子结构。

代码

class TreeNode{
	double val;
	TreeNode lChild;
	TreeNode rChild;
	TreeNode(double val){
		this.val=val;
	}
}

public class Main {
	public static void main(String[] args) {
		//test case
		TreeNode node1=new TreeNode(1);
		TreeNode node2=new TreeNode(2);
		TreeNode node3=new TreeNode(3);
		node1.lChild=node2;
		node1.rChild=node3;
		
		TreeNode node4=new TreeNode(4);
		TreeNode node5=new TreeNode(1);
		TreeNode node6=new TreeNode(2);
		TreeNode node7=new TreeNode(3);
		node4.lChild=node5;
		node5.lChild=node6;
		node5.rChild=node7;
		
		if(hasSubTree(node4,node1)) {
			System.out.print("true");
		}
		else {
			System.out.print("false");
		}
	}
	
	public static boolean hasSubTree(TreeNode root,TreeNode subTreeRoot) {
		 if(subTreeRoot==null) {
			 return true;
		 }
		 if(root==null) {
			 return false;
		 }
		 
		 boolean ans=false;
		 if(equal(root.val,subTreeRoot.val)) {//根节点相同才开始判所在的两棵子树是否相同
			 ans=tree1HasTree2(root,subTreeRoot);
		 }
		 if(!ans) {//
			 ans=hasSubTree(root.lChild,subTreeRoot);
		 }
		 if(!ans) {
			 ans=hasSubTree(root.rChild,subTreeRoot);
		 }
		 return ans;
	}
	
	public static boolean tree1HasTree2(TreeNode root1,TreeNode root2) {
		if(root2==null) {
			return true;
		}
		if(root1==null) {
			return false;
		}
		if(!equal(root1.val,root2.val)) {//递归过程中每次判断节点相同
			return false;
		}
		return tree1HasTree2(root1.lChild,root2.lChild)&&tree1HasTree2(root1.rChild,root2.rChild);
	} 
	
	public static boolean equal(double num1,double num2) {
		return Math.abs(num1-num2)<0.0000001;
	}
}