剑指Offer-18

题目：

输入两棵树 A 和 B，判断 B 是不是 A 的子结构。

实现

//coding = javapublic class Solution18 {    public static boolean hasSubTree(BinaryTreeNode root1,BinaryTreeNode root2){        boolean result = false;        if(root1 == root2){            return true;        }        if(root2==null){            return true;        }        if(root1!=null && root2!=null){            if(root1.value == root2.value){                result = doseTree1HasTree2(root1,root2);            }            if(!result){                result = hasSubTree(root1.left,root2);            }            if(!result){                result = hasSubTree(root1.right,root2);            }        }        return result;    }    public static boolean doseTree1HasTree2(BinaryTreeNode root1,BinaryTreeNode root2){        if(root2==null){            return true;        }        if(root1==null){            return false;        }        if(root1.value!=root2.value){            return false;        }        return doseTree1HasTree2(root1.left,root2.left) && doseTree1HasTree2(root1.right,root2.right);    }    public static void main(String[] args) {        BinaryTreeNode root1 = new BinaryTreeNode();        root1.value = 8;        root1.right = new BinaryTreeNode();        root1.right.value = 7;        root1.left = new BinaryTreeNode();        root1.left.value = 8;        root1.left.left = new BinaryTreeNode();        root1.left.left.value = 9;        root1.left.right = new BinaryTreeNode();        root1.left.right.value = 2;        root1.left.right.left = new BinaryTreeNode();        root1.left.right.left.left = new BinaryTreeNode();        root1.left.right.left.left.value = 4;        root1.left.right.left.right = new BinaryTreeNode();        root1.left.right.left.right.value = 7;        BinaryTreeNode root2 = new BinaryTreeNode();        root2.value = 8;        root2.left = new BinaryTreeNode();        root2.left.value = 9;        root2.right = new BinaryTreeNode();        root2.right.value = 2;        System.out.println(hasSubTree(root1, root2));        System.out.println(hasSubTree(root2, root1));        System.out.println(hasSubTree(root1, root1.left));        System.out.println(hasSubTree(root1, null));        System.out.println(hasSubTree(null, root2));        System.out.println(hasSubTree(null, null));    }}class BinaryTreeNode {    int value;    BinaryTreeNode left;    BinaryTreeNode right;    public BinaryTreeNode() {    }    public BinaryTreeNode(int val) {        this.value = val;    }}