剑指offer:面试题18、树的子结构

题目描述

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

代码示例

public class Offer18 {
    public static void main(String[] args) {
        //构建树1
        TreeNode root1 = new TreeNode(1);
        root1.left = new TreeNode(2);
        root1.right = new TreeNode(3);
        root1.left.left = new TreeNode(4);
        root1.left.right = new TreeNode(5);
        //构建树2
        TreeNode root2 = new TreeNode(1);
        root2.left = new TreeNode(2);
        root2.right = new TreeNode(3);
        //for test
        Offer18 testObj = new Offer18();
        System.out.println(testObj.hasSubTree(root1, root2));

    }

    static class TreeNode {
        int val;
        TreeNode left = null;
        TreeNode right = null;
        TreeNode(int val) {
            this.val = val;
        }
    }

    public boolean hasSubTree(TreeNode root1, TreeNode root2) {
        if (root1 == null || root2 == null) {
            return false;
        }
        return isSubtreeFromRoot(root1,root2) || hasSubTree(root1.left, root2) || hasSubTree(root1.right, root2);
    }
    //从根节点依次向下比较
    private boolean isSubtreeFromRoot(TreeNode root1, TreeNode root2) {
        if (root2 == null)
            return true;//已经递归到root2没有节点了
        if (root1 == null)
            return false;//root2不为null而root1为null所以返回false
        if (root1.val != root2.val)
            return false;
        //递归判定左右节点是否都满足要求
        return isSubtreeFromRoot(root1.left, root2.left)
                && isSubtreeFromRoot(root1.right, root2.right);
    }
}

剑指offer:面试题18、树的子结构

原文：https://www.cnblogs.com/ITxiaolei/p/13166990.html