【剑指26】树的子结构

2019-07-15  本文已影响0人  浅浅星空

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)

分析

public class Solution {
    public  boolean HasSubtree(TreeNode root1, TreeNode root2) {
        if (root1 == null || root2 == null) return false;
        boolean result = false;
        if (root1.val == root2.val) {
            result = isSubtree(root1, root2);
        }
        if (!result) {
            result = HasSubtree(root1.left, root2);
        }
        if (!result) {
            result = HasSubtree(root1.right, root2);
        }
        return result;
    }

    public  boolean isSubtree(TreeNode root1, TreeNode root2) {
        if (root2 == null) return true;
        if (root1 == null) return false;
        if (root1.val != root2.val) return false;
        return isSubtree(root1.left, root2.left) && isSubtree(root1.right, root2.right);
    }
}

class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
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