重拾算法之剑指Offier——树的子结构

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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）

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

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

    }
}

public class Solution {

    public boolean HasSubtree(TreeNode root1, TreeNode root2) {
        boolean flag = false;
        if (root1 != null && root2 != null){
            if (root1.val == root2.val){
                flag = doesTree1haveTree2(root1, root2);
            }
            if(flag == false){
                flag = HasSubtree(root1.left, root2);
            }
            if(flag == false){
                flag = HasSubtree(root1.right, root2);
            }
        }
        return flag;
    }

    public boolean doesTree1haveTree2(TreeNode root1, TreeNode root2){
        if(root2 == null){
            return true;
        }
        if(root1 == null || root1.val != root2.val){
            return false;
        }
        return doesTree1haveTree2(root1.left,root2.left) && doesTree1haveTree2(root1.right,root2.right);
    }
}