劍指 樹的子結構 dfs

啊，太久不寫題自己好菜啊。

題目鏈接

找到兩邊val都相等的root，然後dfs，記得同時滿足左左和右右相等的話用&&

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